---
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.
{ 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.
{ 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=""
```
{ 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.
{ 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: List) {
/* 列表容量 */
fun size(): Int {
return tree.size
}
/* 获取索引为 i 节点的值 */
fun value(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(): List {
val res = ArrayList()
// 直接遍历数组
for (i in 0..) {
// 若为空位，则返回
if (value(i) == null) return
// 前序遍历
if ("pre" == order) res.add(value(i))
dfs(left(i), order, res)
// 中序遍历
if ("in" == order) res.add(value(i))
dfs(right(i), order, res)
// 后序遍历
if ("post" == order) res.add(value(i))
}
/* 前序遍历 */
fun preOrder(): List {
val res = ArrayList()
dfs(0, "pre", res)
return res
}
/* 中序遍历 */
fun inOrder(): List {
val res = ArrayList()
dfs(0, "in", res)
return res
}
/* 后序遍历 */
fun postOrder(): List {
val res = ArrayList()
dfs(0, "post", res)
return res
}
}
```
=== "Ruby"
```ruby title="array_binary_tree.rb"
[class]{ArrayBinaryTree}-[func]{}
```
=== "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.