Table Indexes of the root node and subtrees in preorder and inorder traversals
Table Indexes of the root node and subtrees in pre-order and in-order traversals
+ 
diff --git a/en/docs/chapter_array_and_linkedlist/summary.md b/en/docs/chapter_array_and_linkedlist/summary.md
index 299d5af0e..73b021c7a 100644
--- a/en/docs/chapter_array_and_linkedlist/summary.md
+++ b/en/docs/chapter_array_and_linkedlist/summary.md
@@ -74,7 +74,7 @@ On the other hand, linked lists are primarily necessary for binary trees and gra
**Q**: Does initializing a list `res = [0] * self.size()` result in each element of `res` referencing the same address?
-No. However, this issue arises with two-dimensional arrays, for example, initializing a two-dimensional list `res = [[0] * self.size()]` would reference the same list `[0]` multiple times.
+No. However, this issue arises with two-dimensional arrays, for example, initializing a two-dimensional list `res = [[0]] * self.size()` would reference the same list `[0]` multiple times.
**Q**: In deleting a node, is it necessary to break the reference to its successor node?
diff --git a/en/docs/chapter_backtracking/backtracking_algorithm.md b/en/docs/chapter_backtracking/backtracking_algorithm.md
index eed9cf651..a298ed48c 100644
--- a/en/docs/chapter_backtracking/backtracking_algorithm.md
+++ b/en/docs/chapter_backtracking/backtracking_algorithm.md
@@ -2,19 +2,19 @@
Backtracking algorithm 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 the figure below:
+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 the figure below:
```src
[file]{preorder_traversal_i_compact}-[class]{}-[func]{pre_order}
```
-
+
## Trying and retreating
@@ -425,7 +425,7 @@ As per the requirements, after finding a node with a value of $7$, the search sh
-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.
## Common terminology
diff --git a/en/docs/chapter_divide_and_conquer/build_binary_tree_problem.md b/en/docs/chapter_divide_and_conquer/build_binary_tree_problem.md
index 1d4de018e..0ca7fc009 100644
--- a/en/docs/chapter_divide_and_conquer/build_binary_tree_problem.md
+++ b/en/docs/chapter_divide_and_conquer/build_binary_tree_problem.md
@@ -2,7 +2,7 @@
!!! question
- 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 the figure below).
+ 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 the figure below).
@@ -11,25 +11,25 @@
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.
### 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 the figure below.
-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 ]`.
-
+
### Describing subtree intervals based on variables
@@ -41,7 +41,7 @@ Based on the above division method, **we have now obtained the index intervals o
As shown in the table below, the above variables can represent the index of the root node in `preorder` as well as the index intervals of the subtrees in `inorder`.
-
+