<u>Backtracking algorithm</u> is a method to solve problems by exhaustive search, where the core idea is to start from an initial state and brute force all possible solutions, recording the correct ones until a solution is found or all possible choices are exhausted without finding a solution.
Backtracking typically employs "depth-first search" to traverse the solution space. In the "Binary Tree" chapter, we mentioned that preorder, inorder, and postorder traversals are all depth-first searches. Next, we use preorder traversal to construct a backtracking problem to gradually understand the workings of the backtracking algorithm.
Backtracking typically employs "depth-first search" to traverse the solution space. In the "Binary Tree" chapter, we mentioned that pre-order, in-order, and post-order traversals are all depth-first searches. Next, we use pre-order traversal to construct a backtracking problem to gradually understand the workings of the backtracking algorithm.
!!! question "Example One"
Given a binary tree, search and record all nodes with a value of $7$, please return a list of nodes.
For this problem, we traverse this tree in preorder and check if the current node's value is $7$. If it is, we add the node's value to the result list `res`. The relevant process is shown in Figure 13-1:
For this problem, we traverse this tree in pre-order and check if the current node's value is $7$. If it is, we add the node's value to the result list `res`. The relevant process is shown in Figure 13-1:
=== "Python"
@ -128,9 +128,9 @@ For this problem, we traverse this tree in preorder and check if the current nod
[class]{}-[func]{preOrder}
```
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<palign="center"> Figure 13-1 Searching nodes in preorder traversal </p>
<palign="center"> Figure 13-1 Searching nodes in pre-order traversal </p>
## 13.1.1 Trying and retreating
@ -1110,7 +1110,7 @@ As per the requirements, after finding a node with a value of $7$, the search sh
<palign="center"> Figure 13-4 Comparison of retaining and removing the return in the search process </p>
Compared to the implementation based on preorder traversal, the code implementation based on the backtracking algorithm framework seems verbose, but it has better universality. In fact, **many backtracking problems can be solved within this framework**. We just need to define `state` and `choices` according to the specific problem and implement the methods in the framework.
Compared to the implementation based on pre-order traversal, the code implementation based on the backtracking algorithm framework seems verbose, but it has better universality. In fact, **many backtracking problems can be solved within this framework**. We just need to define `state` and `choices` according to the specific problem and implement the methods in the framework.
Given the preorder traversal `preorder` and inorder traversal `inorder` of a binary tree, construct the binary tree and return the root node of the binary tree. Assume that there are no duplicate values in the nodes of the binary tree (as shown in Figure 12-5).
Given the pre-order traversal `preorder` and in-order traversal `inorder` of a binary tree, construct the binary tree and return the root node of the binary tree. Assume that there are no duplicate values in the nodes of the binary tree (as shown in Figure 12-5).
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@ -17,27 +17,27 @@ comments: true
The original problem of constructing a binary tree from `preorder` and `inorder` is a typical divide and conquer problem.
- **The problem can be decomposed**: From the perspective of divide and conquer, we can divide the original problem into two subproblems: building the left subtree and building the right subtree, plus one operation: initializing the root node. For each subtree (subproblem), we can still use the above division method, dividing it into smaller subtrees (subproblems), until the smallest subproblem (empty subtree) is reached.
- **The subproblems are independent**: The left and right subtrees are independent of each other, with no overlap. When building the left subtree, we only need to focus on the parts of the inorder and preorder traversals that correspond to the left subtree. The same applies to the right subtree.
- **The subproblems are independent**: The left and right subtrees are independent of each other, with no overlap. When building the left subtree, we only need to focus on the parts of the in-order and pre-order traversals that correspond to the left subtree. The same applies to the right subtree.
- **Solutions to subproblems can be combined**: Once the solutions for the left and right subtrees (solutions to subproblems) are obtained, we can link them to the root node to obtain the solution to the original problem.
### 2. How to divide the subtrees
Based on the above analysis, this problem can be solved using divide and conquer, **but how do we use the preorder traversal `preorder` and inorder traversal `inorder` to divide the left and right subtrees?**
Based on the above analysis, this problem can be solved using divide and conquer, **but how do we use the pre-order traversal `preorder` and in-order traversal `inorder` to divide the left and right subtrees?**
By definition, `preorder` and `inorder` can be divided into three parts.
- Preorder traversal: `[ Root | Left Subtree | Right Subtree ]`, for example, the tree in the figure corresponds to `[ 3 | 9 | 2 1 7 ]`.
- Inorder traversal: `[ Left Subtree | Root | Right Subtree ]`, for example, the tree in the figure corresponds to `[ 9 | 3 | 1 2 7 ]`.
- Pre-order traversal: `[ Root | Left Subtree | Right Subtree ]`, for example, the tree in the figure corresponds to `[ 3 | 9 | 2 1 7 ]`.
- In-order traversal: `[ Left Subtree | Root | Right Subtree ]`, for example, the tree in the figure corresponds to `[ 9 | 3 | 1 2 7 ]`.
Using the data in the figure above, we can obtain the division results as shown in Figure 12-6.
1. The first element 3 in the preorder traversal is the value of the root node.
1. The first element 3 in the pre-order traversal is the value of the root node.
2. Find the index of the root node 3 in `inorder`, and use this index to divide `inorder` into `[ 9 | 3 ｜ 1 2 7 ]`.
3. Based on the division results of `inorder`, it is easy to determine the number of nodes in the left and right subtrees as 1 and 3, respectively, thus dividing `preorder` into `[ 3 | 9 | 2 1 7 ]`.
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<palign="center"> Figure 12-6 Dividing the subtrees in preorder and inorder traversals </p>
<palign="center"> Figure 12-6 Dividing the subtrees in pre-order and in-order traversals </p>
### 3. Describing subtree intervals based on variables
@ -49,7 +49,7 @@ Based on the above division method, **we have now obtained the index intervals o
As shown in Table 12-1, the above variables can represent the index of the root node in `preorder` as well as the index intervals of the subtrees in `inorder`.
<palign="center"> Table 12-1 Indexes of the root node and subtrees in preorder and inorder traversals </p>
<palign="center"> Table 12-1 Indexes of the root node and subtrees in pre-order and in-order traversals </p>
- Divide and conquer can solve many algorithm problems and is widely used in data structure and algorithm design, where its presence is ubiquitous.
- Compared to brute force search, adaptive search is more efficient. Search algorithms with a time complexity of $O(\log n)$ are usually based on the divide and conquer strategy.
- Binary search is another typical application of the divide and conquer strategy, which does not include the step of merging the solutions of subproblems. We can implement binary search through recursive divide and conquer.
- In the problem of constructing binary trees, building the tree (original problem) can be divided into building the left and right subtree (subproblems), which can be achieved by partitioning the index intervals of the preorder and inorder traversals.
- In the problem of constructing binary trees, building the tree (original problem) can be divided into building the left and right subtree (subproblems), which can be achieved by partitioning the index intervals of the pre-order and in-order traversals.
- In the Tower of Hanoi problem, a problem of size $n$ can be divided into two subproblems of size $n-1$ and one subproblem of size $1$. By solving these three subproblems in sequence, the original problem is consequently resolved.
@ -460,7 +460,7 @@ To deepen the understanding, it is suggested to combine Figure 9-12 with the cod
Similar to breadth-first traversal, the order of the depth-first traversal sequence is also not unique. Given a certain vertex, exploring in any direction first is possible, that is, the order of adjacent vertices can be arbitrarily shuffled, all being part of depth-first traversal.
Taking tree traversal as an example, "root $\rightarrow$ left $\rightarrow$ right", "left $\rightarrow$ root $\rightarrow$ right", "left $\rightarrow$ right $\rightarrow$ root" correspond to preorder, inorder, and postorder traversals, respectively. They showcase three types of traversal priorities, yet all three are considered depth-first traversal.
Taking tree traversal as an example, "root $\rightarrow$ left $\rightarrow$ right", "left $\rightarrow$ root $\rightarrow$ right", "left $\rightarrow$ right $\rightarrow$ root" correspond to pre-order, in-order, and post-order traversals, respectively. They showcase three types of traversal priorities, yet all three are considered depth-first traversal.
From the perspective of physical structure, a tree is a data structure based on linked lists, hence its traversal method involves accessing nodes one by one through pointers. However, a tree is a non-linear data structure, which makes traversing a tree more complex than traversing a linked list, requiring the assistance of search algorithms to achieve.
Common traversal methods for binary trees include level-order traversal, preorder traversal, inorder traversal, and postorder traversal, among others.
Common traversal methods for binary trees include level-order traversal, pre-order traversal, in-order traversal, and post-order traversal, among others.
## 7.2.1 Level-order traversal
@ -158,15 +158,15 @@ Breadth-first traversal is usually implemented with the help of a "queue". The q
- **Time complexity is $O(n)$**: All nodes are visited once, using $O(n)$ time, where $n$ is the number of nodes.
- **Space complexity is $O(n)$**: In the worst case, i.e., a full binary tree, before traversing to the lowest level, the queue can contain at most $(n + 1) / 2$ nodes at the same time, occupying $O(n)$ space.
## 7.2.2 Preorder, inorder, and postorder traversal
## 7.2.2 Preorder, in-order, and post-order traversal
Correspondingly, preorder, inorder, and postorder traversal all belong to <u>depth-first traversal</u>, also known as <u>depth-first search (DFS)</u>, which embodies a "proceed to the end first, then backtrack and continue" traversal method.
Correspondingly, pre-order, in-order, and post-order traversal all belong to <u>depth-first traversal</u>, also known as <u>depth-first search (DFS)</u>, which embodies a "proceed to the end first, then backtrack and continue" traversal method.
Figure 7-10 shows the working principle of performing a depth-first traversal on a binary tree. **Depth-first traversal is like walking around the perimeter of the entire binary tree**, encountering three positions at each node, corresponding to preorder traversal, inorder traversal, and postorder traversal.
Figure 7-10 shows the working principle of performing a depth-first traversal on a binary tree. **Depth-first traversal is like walking around the perimeter of the entire binary tree**, encountering three positions at each node, corresponding to pre-order traversal, in-order traversal, and post-order traversal.
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<palign="center"> Figure 7-10 Preorder, inorder, and postorder traversal of a binary search tree </p>
<palign="center"> Figure 7-10 Preorder, in-order, and post-order traversal of a binary search tree </p>
### 1. Code implementation
@ -385,13 +385,13 @@ Depth-first search is usually implemented based on recursion:
Depth-first search can also be implemented based on iteration, interested readers can study this on their own.
Figure 7-11 shows the recursive process of preorder traversal of a binary tree, which can be divided into two opposite parts: "recursion" and "return".
Figure 7-11 shows the recursive process of pre-order traversal of a binary tree, which can be divided into two opposite parts: "recursion" and "return".
1. "Recursion" means starting a new method, the program accesses the next node in this process.
2. "Return" means the function returns, indicating the current node has been fully accessed.
=== "<1>"
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krahets
6 months ago