--- comments: true --- # 7.3   Array representation of binary trees Under the linked list representation, the storage unit of a binary tree is a node `TreeNode`, with nodes connected by pointers. The basic operations of binary trees under the linked list representation were introduced in the previous section. So, can we use an array to represent a binary tree? The answer is yes. ## 7.3.1   Representing perfect binary trees Let's analyze a simple case first. Given a perfect binary tree, we store all nodes in an array according to the order of level-order traversal, where each node corresponds to a unique array index. Based on the characteristics of level-order traversal, we can deduce a "mapping formula" between the index of a parent node and its children: **If a node's index is $i$, then the index of its left child is $2i + 1$ and the right child is $2i + 2$**. The Figure 7-12 shows the mapping relationship between the indices of various nodes. ![Array representation of a perfect binary tree](array_representation_of_tree.assets/array_representation_binary_tree.png){ class="animation-figure" }

Figure 7-12   Array representation of a perfect binary tree

**The mapping formula plays a role similar to the node references (pointers) in linked lists**. Given any node in the array, we can access its left (right) child node using the mapping formula. ## 7.3.2   Representing any binary tree Perfect binary trees are a special case; there are often many `None` values in the middle levels of a binary tree. Since the sequence of level-order traversal does not include these `None` values, we cannot solely rely on this sequence to deduce the number and distribution of `None` values. **This means that multiple binary tree structures can match the same level-order traversal sequence**. As shown in the Figure 7-13 , given a non-perfect binary tree, the above method of array representation fails. ![Level-order traversal sequence corresponds to multiple binary tree possibilities](array_representation_of_tree.assets/array_representation_without_empty.png){ class="animation-figure" }

Figure 7-13   Level-order traversal sequence corresponds to multiple binary tree possibilities

To solve this problem, **we can consider explicitly writing out all `None` values in the level-order traversal sequence**. As shown in the following figure, after this treatment, the level-order traversal sequence can uniquely represent a binary tree. Example code is as follows: === "Python" ```python title="" # Array representation of a binary tree # Using None to represent empty slots tree = [1, 2, 3, 4, None, 6, 7, 8, 9, None, None, 12, None, None, 15] ``` === "C++" ```cpp title="" /* Array representation of a binary tree */ // Using the maximum integer value INT_MAX to mark empty slots vector tree = {1, 2, 3, 4, INT_MAX, 6, 7, 8, 9, INT_MAX, INT_MAX, 12, INT_MAX, INT_MAX, 15}; ``` === "Java" ```java title="" /* Array representation of a binary tree */ // Using the Integer wrapper class allows for using null to mark empty slots Integer[] tree = { 1, 2, 3, 4, null, 6, 7, 8, 9, null, null, 12, null, null, 15 }; ``` === "C#" ```csharp title="" /* Array representation of a binary tree */ // Using nullable int (int?) allows for using null to mark empty slots int?[] tree = [1, 2, 3, 4, null, 6, 7, 8, 9, null, null, 12, null, null, 15]; ``` === "Go" ```go title="" /* Array representation of a binary tree */ // Using an any type slice, allowing for nil to mark empty slots tree := []any{1, 2, 3, 4, nil, 6, 7, 8, 9, nil, nil, 12, nil, nil, 15} ``` === "Swift" ```swift title="" /* Array representation of a binary tree */ // Using optional Int (Int?) allows for using nil to mark empty slots let tree: [Int?] = [1, 2, 3, 4, nil, 6, 7, 8, 9, nil, nil, 12, nil, nil, 15] ``` === "JS" ```javascript title="" /* Array representation of a binary tree */ // Using null to represent empty slots let tree = [1, 2, 3, 4, null, 6, 7, 8, 9, null, null, 12, null, null, 15]; ``` === "TS" ```typescript title="" /* Array representation of a binary tree */ // Using null to represent empty slots let tree: (number | null)[] = [1, 2, 3, 4, null, 6, 7, 8, 9, null, null, 12, null, null, 15]; ``` === "Dart" ```dart title="" /* Array representation of a binary tree */ // Using nullable int (int?) allows for using null to mark empty slots List tree = [1, 2, 3, 4, null, 6, 7, 8, 9, null, null, 12, null, null, 15]; ``` === "Rust" ```rust title="" /* Array representation of a binary tree */ // Using None to mark empty slots let tree = [Some(1), Some(2), Some(3), Some(4), None, Some(6), Some(7), Some(8), Some(9), None, None, Some(12), None, None, Some(15)]; ``` === "C" ```c title="" /* Array representation of a binary tree */ // Using the maximum int value to mark empty slots, therefore, node values must not be INT_MAX int tree[] = {1, 2, 3, 4, INT_MAX, 6, 7, 8, 9, INT_MAX, INT_MAX, 12, INT_MAX, INT_MAX, 15}; ``` === "Kotlin" ```kotlin title="" /* Array representation of a binary tree */ // Using null to represent empty slots val tree = mutableListOf( 1, 2, 3, 4, null, 6, 7, 8, 9, null, null, 12, null, null, 15 ) ``` === "Ruby" ```ruby title="" ``` === "Zig" ```zig title="" ``` ![Array representation of any type of binary tree](array_representation_of_tree.assets/array_representation_with_empty.png){ class="animation-figure" }

Figure 7-14   Array representation of any type of binary tree

It's worth noting that **complete binary trees are very suitable for array representation**. Recalling the definition of a complete binary tree, `None` appears only at the bottom level and towards the right, **meaning all `None` values definitely appear at the end of the level-order traversal sequence**. This means that when using an array to represent a complete binary tree, it's possible to omit storing all `None` values, which is very convenient. The Figure 7-15 gives an example. ![Array representation of a complete binary tree](array_representation_of_tree.assets/array_representation_complete_binary_tree.png){ class="animation-figure" }

Figure 7-15   Array representation of a complete binary tree

The following code implements a binary tree based on array representation, including the following operations: - Given a node, obtain its value, left (right) child node, and parent node. - Obtain the preorder, inorder, postorder, and level-order traversal sequences. === "Python" ```python title="array_binary_tree.py" class ArrayBinaryTree: """数组表示下的二叉树类""" def __init__(self, arr: list[int | None]): """构造方法""" self._tree = list(arr) def size(self): """列表容量""" return len(self._tree) def val(self, i: int) -> int: """获取索引为 i 节点的值""" # 若索引越界,则返回 None ,代表空位 if i < 0 or i >= self.size(): return None return self._tree[i] def left(self, i: int) -> int | None: """获取索引为 i 节点的左子节点的索引""" return 2 * i + 1 def right(self, i: int) -> int | None: """获取索引为 i 节点的右子节点的索引""" return 2 * i + 2 def parent(self, i: int) -> int | None: """获取索引为 i 节点的父节点的索引""" return (i - 1) // 2 def level_order(self) -> list[int]: """层序遍历""" self.res = [] # 直接遍历数组 for i in range(self.size()): if self.val(i) is not None: self.res.append(self.val(i)) return self.res def dfs(self, i: int, order: str): """深度优先遍历""" if self.val(i) is None: return # 前序遍历 if order == "pre": self.res.append(self.val(i)) self.dfs(self.left(i), order) # 中序遍历 if order == "in": self.res.append(self.val(i)) self.dfs(self.right(i), order) # 后序遍历 if order == "post": self.res.append(self.val(i)) def pre_order(self) -> list[int]: """前序遍历""" self.res = [] self.dfs(0, order="pre") return self.res def in_order(self) -> list[int]: """中序遍历""" self.res = [] self.dfs(0, order="in") return self.res def post_order(self) -> list[int]: """后序遍历""" self.res = [] self.dfs(0, order="post") return self.res ``` === "C++" ```cpp title="array_binary_tree.cpp" /* 数组表示下的二叉树类 */ class ArrayBinaryTree { public: /* 构造方法 */ ArrayBinaryTree(vector arr) { tree = arr; } /* 列表容量 */ int size() { return tree.size(); } /* 获取索引为 i 节点的值 */ int val(int i) { // 若索引越界,则返回 INT_MAX ,代表空位 if (i < 0 || i >= size()) return INT_MAX; return tree[i]; } /* 获取索引为 i 节点的左子节点的索引 */ int left(int i) { return 2 * i + 1; } /* 获取索引为 i 节点的右子节点的索引 */ int right(int i) { return 2 * i + 2; } /* 获取索引为 i 节点的父节点的索引 */ int parent(int i) { return (i - 1) / 2; } /* 层序遍历 */ vector levelOrder() { vector res; // 直接遍历数组 for (int i = 0; i < size(); i++) { if (val(i) != INT_MAX) res.push_back(val(i)); } return res; } /* 前序遍历 */ vector preOrder() { vector res; dfs(0, "pre", res); return res; } /* 中序遍历 */ vector inOrder() { vector res; dfs(0, "in", res); return res; } /* 后序遍历 */ vector postOrder() { vector res; dfs(0, "post", res); return res; } private: vector tree; /* 深度优先遍历 */ void dfs(int i, string order, vector &res) { // 若为空位,则返回 if (val(i) == INT_MAX) return; // 前序遍历 if (order == "pre") res.push_back(val(i)); dfs(left(i), order, res); // 中序遍历 if (order == "in") res.push_back(val(i)); dfs(right(i), order, res); // 后序遍历 if (order == "post") res.push_back(val(i)); } }; ``` === "Java" ```java title="array_binary_tree.java" /* 数组表示下的二叉树类 */ class ArrayBinaryTree { private List tree; /* 构造方法 */ public ArrayBinaryTree(List arr) { tree = new ArrayList<>(arr); } /* 列表容量 */ public int size() { return tree.size(); } /* 获取索引为 i 节点的值 */ public Integer val(int i) { // 若索引越界,则返回 null ,代表空位 if (i < 0 || i >= size()) return null; return tree.get(i); } /* 获取索引为 i 节点的左子节点的索引 */ public Integer left(int i) { return 2 * i + 1; } /* 获取索引为 i 节点的右子节点的索引 */ public Integer right(int i) { return 2 * i + 2; } /* 获取索引为 i 节点的父节点的索引 */ public Integer parent(int i) { return (i - 1) / 2; } /* 层序遍历 */ public List levelOrder() { List res = new ArrayList<>(); // 直接遍历数组 for (int i = 0; i < size(); i++) { if (val(i) != null) res.add(val(i)); } return res; } /* 深度优先遍历 */ private void dfs(Integer i, String order, List res) { // 若为空位,则返回 if (val(i) == null) return; // 前序遍历 if ("pre".equals(order)) res.add(val(i)); dfs(left(i), order, res); // 中序遍历 if ("in".equals(order)) res.add(val(i)); dfs(right(i), order, res); // 后序遍历 if ("post".equals(order)) res.add(val(i)); } /* 前序遍历 */ public List preOrder() { List res = new ArrayList<>(); dfs(0, "pre", res); return res; } /* 中序遍历 */ public List inOrder() { List res = new ArrayList<>(); dfs(0, "in", res); return res; } /* 后序遍历 */ public List postOrder() { List res = new ArrayList<>(); dfs(0, "post", res); return res; } } ``` === "C#" ```csharp title="array_binary_tree.cs" /* 数组表示下的二叉树类 */ class ArrayBinaryTree(List arr) { List tree = new(arr); /* 列表容量 */ public int Size() { return tree.Count; } /* 获取索引为 i 节点的值 */ public int? Val(int i) { // 若索引越界,则返回 null ,代表空位 if (i < 0 || i >= Size()) return null; return tree[i]; } /* 获取索引为 i 节点的左子节点的索引 */ public int Left(int i) { return 2 * i + 1; } /* 获取索引为 i 节点的右子节点的索引 */ public int Right(int i) { return 2 * i + 2; } /* 获取索引为 i 节点的父节点的索引 */ public int Parent(int i) { return (i - 1) / 2; } /* 层序遍历 */ public List LevelOrder() { List res = []; // 直接遍历数组 for (int i = 0; i < Size(); i++) { if (Val(i).HasValue) res.Add(Val(i)!.Value); } return res; } /* 深度优先遍历 */ void DFS(int i, string order, List res) { // 若为空位,则返回 if (!Val(i).HasValue) return; // 前序遍历 if (order == "pre") res.Add(Val(i)!.Value); DFS(Left(i), order, res); // 中序遍历 if (order == "in") res.Add(Val(i)!.Value); DFS(Right(i), order, res); // 后序遍历 if (order == "post") res.Add(Val(i)!.Value); } /* 前序遍历 */ public List PreOrder() { List res = []; DFS(0, "pre", res); return res; } /* 中序遍历 */ public List InOrder() { List res = []; DFS(0, "in", res); return res; } /* 后序遍历 */ public List PostOrder() { List res = []; DFS(0, "post", res); return res; } } ``` === "Go" ```go title="array_binary_tree.go" /* 数组表示下的二叉树类 */ type arrayBinaryTree struct { tree []any } /* 构造方法 */ func newArrayBinaryTree(arr []any) *arrayBinaryTree { return &arrayBinaryTree{ tree: arr, } } /* 列表容量 */ func (abt *arrayBinaryTree) size() int { return len(abt.tree) } /* 获取索引为 i 节点的值 */ func (abt *arrayBinaryTree) val(i int) any { // 若索引越界,则返回 null ,代表空位 if i < 0 || i >= abt.size() { return nil } return abt.tree[i] } /* 获取索引为 i 节点的左子节点的索引 */ func (abt *arrayBinaryTree) left(i int) int { return 2*i + 1 } /* 获取索引为 i 节点的右子节点的索引 */ func (abt *arrayBinaryTree) right(i int) int { return 2*i + 2 } /* 获取索引为 i 节点的父节点的索引 */ func (abt *arrayBinaryTree) parent(i int) int { return (i - 1) / 2 } /* 层序遍历 */ func (abt *arrayBinaryTree) levelOrder() []any { var res []any // 直接遍历数组 for i := 0; i < abt.size(); i++ { if abt.val(i) != nil { res = append(res, abt.val(i)) } } return res } /* 深度优先遍历 */ func (abt *arrayBinaryTree) dfs(i int, order string, res *[]any) { // 若为空位,则返回 if abt.val(i) == nil { return } // 前序遍历 if order == "pre" { *res = append(*res, abt.val(i)) } abt.dfs(abt.left(i), order, res) // 中序遍历 if order == "in" { *res = append(*res, abt.val(i)) } abt.dfs(abt.right(i), order, res) // 后序遍历 if order == "post" { *res = append(*res, abt.val(i)) } } /* 前序遍历 */ func (abt *arrayBinaryTree) preOrder() []any { var res []any abt.dfs(0, "pre", &res) return res } /* 中序遍历 */ func (abt *arrayBinaryTree) inOrder() []any { var res []any abt.dfs(0, "in", &res) return res } /* 后序遍历 */ func (abt *arrayBinaryTree) postOrder() []any { var res []any abt.dfs(0, "post", &res) return res } ``` === "Swift" ```swift title="array_binary_tree.swift" /* 数组表示下的二叉树类 */ class ArrayBinaryTree { private var tree: [Int?] /* 构造方法 */ init(arr: [Int?]) { tree = arr } /* 列表容量 */ func size() -> Int { tree.count } /* 获取索引为 i 节点的值 */ func val(i: Int) -> Int? { // 若索引越界,则返回 null ,代表空位 if i < 0 || i >= size() { return nil } return tree[i] } /* 获取索引为 i 节点的左子节点的索引 */ func left(i: Int) -> Int { 2 * i + 1 } /* 获取索引为 i 节点的右子节点的索引 */ func right(i: Int) -> Int { 2 * i + 2 } /* 获取索引为 i 节点的父节点的索引 */ func parent(i: Int) -> Int { (i - 1) / 2 } /* 层序遍历 */ func levelOrder() -> [Int] { var res: [Int] = [] // 直接遍历数组 for i in 0 ..< size() { if let val = val(i: i) { res.append(val) } } return res } /* 深度优先遍历 */ private func dfs(i: Int, order: String, res: inout [Int]) { // 若为空位,则返回 guard let val = val(i: i) else { return } // 前序遍历 if order == "pre" { res.append(val) } dfs(i: left(i: i), order: order, res: &res) // 中序遍历 if order == "in" { res.append(val) } dfs(i: right(i: i), order: order, res: &res) // 后序遍历 if order == "post" { res.append(val) } } /* 前序遍历 */ func preOrder() -> [Int] { var res: [Int] = [] dfs(i: 0, order: "pre", res: &res) return res } /* 中序遍历 */ func inOrder() -> [Int] { var res: [Int] = [] dfs(i: 0, order: "in", res: &res) return res } /* 后序遍历 */ func postOrder() -> [Int] { var res: [Int] = [] dfs(i: 0, order: "post", res: &res) return res } } ``` === "JS" ```javascript title="array_binary_tree.js" /* 数组表示下的二叉树类 */ class ArrayBinaryTree { #tree; /* 构造方法 */ constructor(arr) { this.#tree = arr; } /* 列表容量 */ size() { return this.#tree.length; } /* 获取索引为 i 节点的值 */ val(i) { // 若索引越界,则返回 null ,代表空位 if (i < 0 || i >= this.size()) return null; return this.#tree[i]; } /* 获取索引为 i 节点的左子节点的索引 */ left(i) { return 2 * i + 1; } /* 获取索引为 i 节点的右子节点的索引 */ right(i) { return 2 * i + 2; } /* 获取索引为 i 节点的父节点的索引 */ parent(i) { return Math.floor((i - 1) / 2); // 向下整除 } /* 层序遍历 */ levelOrder() { let res = []; // 直接遍历数组 for (let i = 0; i < this.size(); i++) { if (this.val(i) !== null) res.push(this.val(i)); } return res; } /* 深度优先遍历 */ #dfs(i, order, res) { // 若为空位,则返回 if (this.val(i) === null) return; // 前序遍历 if (order === 'pre') res.push(this.val(i)); this.#dfs(this.left(i), order, res); // 中序遍历 if (order === 'in') res.push(this.val(i)); this.#dfs(this.right(i), order, res); // 后序遍历 if (order === 'post') res.push(this.val(i)); } /* 前序遍历 */ preOrder() { const res = []; this.#dfs(0, 'pre', res); return res; } /* 中序遍历 */ inOrder() { const res = []; this.#dfs(0, 'in', res); return res; } /* 后序遍历 */ postOrder() { const res = []; this.#dfs(0, 'post', res); return res; } } ``` === "TS" ```typescript title="array_binary_tree.ts" /* 数组表示下的二叉树类 */ class ArrayBinaryTree { #tree: (number | null)[]; /* 构造方法 */ constructor(arr: (number | null)[]) { this.#tree = arr; } /* 列表容量 */ size(): number { return this.#tree.length; } /* 获取索引为 i 节点的值 */ val(i: number): number | null { // 若索引越界,则返回 null ,代表空位 if (i < 0 || i >= this.size()) return null; return this.#tree[i]; } /* 获取索引为 i 节点的左子节点的索引 */ left(i: number): number { return 2 * i + 1; } /* 获取索引为 i 节点的右子节点的索引 */ right(i: number): number { return 2 * i + 2; } /* 获取索引为 i 节点的父节点的索引 */ parent(i: number): number { return Math.floor((i - 1) / 2); // 向下整除 } /* 层序遍历 */ levelOrder(): number[] { let res = []; // 直接遍历数组 for (let i = 0; i < this.size(); i++) { if (this.val(i) !== null) res.push(this.val(i)); } return res; } /* 深度优先遍历 */ #dfs(i: number, order: Order, res: (number | null)[]): void { // 若为空位,则返回 if (this.val(i) === null) return; // 前序遍历 if (order === 'pre') res.push(this.val(i)); this.#dfs(this.left(i), order, res); // 中序遍历 if (order === 'in') res.push(this.val(i)); this.#dfs(this.right(i), order, res); // 后序遍历 if (order === 'post') res.push(this.val(i)); } /* 前序遍历 */ preOrder(): (number | null)[] { const res = []; this.#dfs(0, 'pre', res); return res; } /* 中序遍历 */ inOrder(): (number | null)[] { const res = []; this.#dfs(0, 'in', res); return res; } /* 后序遍历 */ postOrder(): (number | null)[] { const res = []; this.#dfs(0, 'post', res); return res; } } ``` === "Dart" ```dart title="array_binary_tree.dart" /* 数组表示下的二叉树类 */ class ArrayBinaryTree { late List _tree; /* 构造方法 */ ArrayBinaryTree(this._tree); /* 列表容量 */ int size() { return _tree.length; } /* 获取索引为 i 节点的值 */ int? val(int i) { // 若索引越界,则返回 null ,代表空位 if (i < 0 || i >= size()) { return null; } return _tree[i]; } /* 获取索引为 i 节点的左子节点的索引 */ int? left(int i) { return 2 * i + 1; } /* 获取索引为 i 节点的右子节点的索引 */ int? right(int i) { return 2 * i + 2; } /* 获取索引为 i 节点的父节点的索引 */ int? parent(int i) { return (i - 1) ~/ 2; } /* 层序遍历 */ List levelOrder() { List res = []; for (int i = 0; i < size(); i++) { if (val(i) != null) { res.add(val(i)!); } } return res; } /* 深度优先遍历 */ void dfs(int i, String order, List res) { // 若为空位,则返回 if (val(i) == null) { return; } // 前序遍历 if (order == 'pre') { res.add(val(i)); } dfs(left(i)!, order, res); // 中序遍历 if (order == 'in') { res.add(val(i)); } dfs(right(i)!, order, res); // 后序遍历 if (order == 'post') { res.add(val(i)); } } /* 前序遍历 */ List preOrder() { List res = []; dfs(0, 'pre', res); return res; } /* 中序遍历 */ List inOrder() { List res = []; dfs(0, 'in', res); return res; } /* 后序遍历 */ List postOrder() { List res = []; dfs(0, 'post', res); return res; } } ``` === "Rust" ```rust title="array_binary_tree.rs" /* 数组表示下的二叉树类 */ struct ArrayBinaryTree { tree: Vec>, } impl ArrayBinaryTree { /* 构造方法 */ fn new(arr: Vec>) -> Self { Self { tree: arr } } /* 列表容量 */ fn size(&self) -> i32 { self.tree.len() as i32 } /* 获取索引为 i 节点的值 */ fn val(&self, i: i32) -> Option { // 若索引越界,则返回 None ,代表空位 if i < 0 || i >= self.size() { None } else { self.tree[i as usize] } } /* 获取索引为 i 节点的左子节点的索引 */ fn left(&self, i: i32) -> i32 { 2 * i + 1 } /* 获取索引为 i 节点的右子节点的索引 */ fn right(&self, i: i32) -> i32 { 2 * i + 2 } /* 获取索引为 i 节点的父节点的索引 */ fn parent(&self, i: i32) -> i32 { (i - 1) / 2 } /* 层序遍历 */ fn level_order(&self) -> Vec { let mut res = vec![]; // 直接遍历数组 for i in 0..self.size() { if let Some(val) = self.val(i) { res.push(val) } } res } /* 深度优先遍历 */ fn dfs(&self, i: i32, order: &str, res: &mut Vec) { if self.val(i).is_none() { return; } let val = self.val(i).unwrap(); // 前序遍历 if order == "pre" { res.push(val); } self.dfs(self.left(i), order, res); // 中序遍历 if order == "in" { res.push(val); } self.dfs(self.right(i), order, res); // 后序遍历 if order == "post" { res.push(val); } } /* 前序遍历 */ fn pre_order(&self) -> Vec { let mut res = vec![]; self.dfs(0, "pre", &mut res); res } /* 中序遍历 */ fn in_order(&self) -> Vec { let mut res = vec![]; self.dfs(0, "in", &mut res); res } /* 后序遍历 */ fn post_order(&self) -> Vec { let mut res = vec![]; self.dfs(0, "post", &mut res); res } } ``` === "C" ```c title="array_binary_tree.c" /* 数组表示下的二叉树结构体 */ typedef struct { int *tree; int size; } ArrayBinaryTree; /* 构造函数 */ ArrayBinaryTree *newArrayBinaryTree(int *arr, int arrSize) { ArrayBinaryTree *abt = (ArrayBinaryTree *)malloc(sizeof(ArrayBinaryTree)); abt->tree = malloc(sizeof(int) * arrSize); memcpy(abt->tree, arr, sizeof(int) * arrSize); abt->size = arrSize; return abt; } /* 析构函数 */ void delArrayBinaryTree(ArrayBinaryTree *abt) { free(abt->tree); free(abt); } /* 列表容量 */ int size(ArrayBinaryTree *abt) { return abt->size; } /* 获取索引为 i 节点的值 */ int val(ArrayBinaryTree *abt, int i) { // 若索引越界,则返回 INT_MAX ,代表空位 if (i < 0 || i >= size(abt)) return INT_MAX; return abt->tree[i]; } /* 层序遍历 */ int *levelOrder(ArrayBinaryTree *abt, int *returnSize) { int *res = (int *)malloc(sizeof(int) * size(abt)); int index = 0; // 直接遍历数组 for (int i = 0; i < size(abt); i++) { if (val(abt, i) != INT_MAX) res[index++] = val(abt, i); } *returnSize = index; return res; } /* 深度优先遍历 */ void dfs(ArrayBinaryTree *abt, int i, char *order, int *res, int *index) { // 若为空位,则返回 if (val(abt, i) == INT_MAX) return; // 前序遍历 if (strcmp(order, "pre") == 0) res[(*index)++] = val(abt, i); dfs(abt, left(i), order, res, index); // 中序遍历 if (strcmp(order, "in") == 0) res[(*index)++] = val(abt, i); dfs(abt, right(i), order, res, index); // 后序遍历 if (strcmp(order, "post") == 0) res[(*index)++] = val(abt, i); } /* 前序遍历 */ int *preOrder(ArrayBinaryTree *abt, int *returnSize) { int *res = (int *)malloc(sizeof(int) * size(abt)); int index = 0; dfs(abt, 0, "pre", res, &index); *returnSize = index; return res; } /* 中序遍历 */ int *inOrder(ArrayBinaryTree *abt, int *returnSize) { int *res = (int *)malloc(sizeof(int) * size(abt)); int index = 0; dfs(abt, 0, "in", res, &index); *returnSize = index; return res; } /* 后序遍历 */ int *postOrder(ArrayBinaryTree *abt, int *returnSize) { int *res = (int *)malloc(sizeof(int) * size(abt)); int index = 0; dfs(abt, 0, "post", res, &index); *returnSize = index; return res; } ``` === "Kotlin" ```kotlin title="array_binary_tree.kt" /* 数组表示下的二叉树类 */ class ArrayBinaryTree(val tree: MutableList) { /* 列表容量 */ fun size(): Int { return tree.size } /* 获取索引为 i 节点的值 */ fun _val(i: Int): Int? { // 若索引越界,则返回 null ,代表空位 if (i < 0 || i >= size()) return null return tree[i] } /* 获取索引为 i 节点的左子节点的索引 */ fun left(i: Int): Int { return 2 * i + 1 } /* 获取索引为 i 节点的右子节点的索引 */ fun right(i: Int): Int { return 2 * i + 2 } /* 获取索引为 i 节点的父节点的索引 */ fun parent(i: Int): Int { return (i - 1) / 2 } /* 层序遍历 */ fun levelOrder(): MutableList { val res = mutableListOf() // 直接遍历数组 for (i in 0..) { // 若为空位,则返回 if (_val(i) == null) return // 前序遍历 if ("pre" == order) res.add(_val(i)) dfs(left(i), order, res) // 中序遍历 if ("in" == order) res.add(_val(i)) dfs(right(i), order, res) // 后序遍历 if ("post" == order) res.add(_val(i)) } /* 前序遍历 */ fun preOrder(): MutableList { val res = mutableListOf() dfs(0, "pre", res) return res } /* 中序遍历 */ fun inOrder(): MutableList { val res = mutableListOf() dfs(0, "in", res) return res } /* 后序遍历 */ fun postOrder(): MutableList { val res = mutableListOf() dfs(0, "post", res) return res } } ``` === "Ruby" ```ruby title="array_binary_tree.rb" ### 数组表示下的二叉树类 ### class ArrayBinaryTree ### 构造方法 ### def initialize(arr) @tree = arr.to_a end ### 列表容量 ### def size @tree.length end ### 获取索引为 i 节点的值 ### def val(i) # 若索引越界,则返回 nil ,代表空位 return if i < 0 || i >= size @tree[i] end ### 获取索引为 i 节点的左子节点的索引 ### def left(i) 2 * i + 1 end ### 获取索引为 i 节点的右子节点的索引 ### def right(i) 2 * i + 2 end ### 获取索引为 i 节点的父节点的索引 ### def parent(i) (i - 1) / 2 end ### 层序遍历 ### def level_order @res = [] # 直接遍历数组 for i in 0...size @res << val(i) unless val(i).nil? end @res end ### 深度优先遍历 ### def dfs(i, order) return if val(i).nil? # 前序遍历 @res << val(i) if order == :pre dfs(left(i), order) # 中序遍历 @res << val(i) if order == :in dfs(right(i), order) # 后序遍历 @res << val(i) if order == :post end ### 前序遍历 ### def pre_order @res = [] dfs(0, :pre) @res end ### 中序遍历 ### def in_order @res = [] dfs(0, :in) @res end ### 后序遍历 ### def post_order @res = [] dfs(0, :post) @res end end ``` === "Zig" ```zig title="array_binary_tree.zig" [class]{ArrayBinaryTree}-[func]{} ``` ??? pythontutor "Code Visualization"
## 7.3.3   Advantages and limitations The array representation of binary trees has the following advantages: - Arrays are stored in contiguous memory spaces, which is cache-friendly and allows for faster access and traversal. - It does not require storing pointers, which saves space. - It allows random access to nodes. However, the array representation also has some limitations: - Array storage requires contiguous memory space, so it is not suitable for storing trees with a large amount of data. - Adding or deleting nodes requires array insertion and deletion operations, which are less efficient. - When there are many `None` values in the binary tree, the proportion of node data contained in the array is low, leading to lower space utilization.