【力扣 - 二叉树的中序遍历】-Toy模板网

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题目描述

给定一个二叉树的根节点 root ，返回它的中序遍历。
【力扣 - 二叉树的中序遍历】,C语言学习报告,leetcode,算法

提示：

树中节点数目在范围 [0, 100] 内

-100 <= Node.val <= 100

方法一：递归

思路与算法

首先我们需要了解什么是二叉树的中序遍历：按照访问左子树——根节点——右子树的方式遍历这棵树，而在访问左子树或者右子树的时候我们按照同样的方式遍历，直到遍历完整棵树。因此整个遍历过程天然具有递归的性质，我们可以直接用递归函数来模拟这一过程。

定义 inorder(root) 表示当前遍历到 root 节点的答案，那么按照定义，我们只要递归调用 inorder(root.left) 来遍历 root 节点的左子树，然后将 root 节点的值加入答案，再递归调用inorder(root.right) 来遍历 root 节点的右子树即可，递归终止的条件为碰到空节点。

代码

/**
 * Definition for a binary tree node.
 */
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};
/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
 /*
  * The  inorder  function performs an inorder traversal of a binary tree recursively. 
  * It stores the values of the nodes in the result array  res  and increments the size  resSize  accordingly. 
 */
 /*
  * The function "inorder" in the provided code is a recursive function. 
  * When a function calls itself inside its own definition, it is known as recursion. 
  * In this case, the "inorder" function is designed to perform an inorder traversal of a binary tree. 
  * 1. The "inorder" function is called with the root node of the binary tree.
  * 2. Inside the function, it first checks if the current node is NULL. If it is NULL, the function returns and the recursion stops.
  * 3. If the current node is not NULL, the function recursively calls itself for the left child of the current node (root->left). This step continues until it reaches a NULL node (i.e., the left subtree is fully traversed).
  * 4. After traversing the left subtree, the function stores the value of the current node in the result array and increments the size of the result array.
  * 5. Finally, the function recursively calls itself for the right child of the current node (root->right) to traverse the right subtree.
  * This recursive process repeats for each node in the binary tree, 
  * effectively performing an inorder traversal by visiting the nodes in the order of left subtree - current node - right subtree. 
  * Each recursive call maintains its own set of variables and execution context, 
  * allowing the function to traverse the entire tree in an ordered manner.
  */ 
void inorder(struct TreeNode* root, int* res, int* resSize) {
    // Check if the current node is NULL
    if (!root) {
        return;  // Return if the current node is NULL
    }
    
    // Traverse the left subtree in inorder
    inorder(root->left, res, resSize);
    
    // Store the value of the current node in the result array and increment the size
    res[(*resSize)++] = root->val;
    
    // Traverse the right subtree in inorder
    inorder(root->right, res, resSize);
    /*
     * `res[(*resSize)++] = root->val;`  is not needed here,
     * because the inorder traversal of a binary tree is structured in such a way that after traversing the left subtree and the current node, 
     * the traversal of the right subtree will naturally continue the process of storing the values in the correct order in the result array  `res` .
     * In an inorder traversal, the sequence of operations ensures that the left subtree is fully explored before visiting the current node, 
     * and then the right subtree is explored after the current node. 
     * Therefore, by the time the function returns from the recursive call  `inorder(root->right, res, resSize);` , 
     * the right subtree has been traversed and the values have been stored in the result array in the correct order relative to the current node.
     * Including  `res[(*resSize)++] = root->val;`  after the right subtree traversal would result in duplicating the value of the current node in the result array, 
     * which is unnecessary and would disrupt the correct inorder traversal sequence.
     */
}
/*
 * The  inorderTraversal  function initializes the result array, 
 * calls the  inorder  function to perform the traversal, and then returns the result array. 
 */ 
int* inorderTraversal(struct TreeNode* root, int* returnSize) {
    // Allocate memory for the result array
    // Create an integer array of size 501 dynamically on the heap and assigning the address of the first element of the array to the pointer variable  res .
    int* res = malloc(sizeof(int) * 501);
    
    // Initialize the return size to 0
    *returnSize = 0;
    
    // Perform inorder traversal starting from the root node
    inorder(root, res, returnSize);
    
    // Return the result array containing the inorder traversal of the binary tree
    return res;
}

复杂度分析

时间复杂度：O(n)，其中 n 为二叉树节点的个数。二叉树的遍历中每个节点会被访问一次且只会被访问一次。
空间复杂度：O(n)。空间复杂度取决于递归的栈深度，而栈深度在二叉树为一条链的情况下会达到 O(n)的级别。

方法二：迭代

思路与算法

方法一的递归函数我们也可以用迭代的方式实现，两种方式是等价的，区别在于递归的时候隐式地维护了一个栈，而我们在迭代的时候需要显式地将这个栈模拟出来，其他都相同。
【力扣 - 二叉树的中序遍历】,C语言学习报告,leetcode,算法

代码

/**
 * Definition for a binary tree node.
 */
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};
/**
 * An iterative version of the inorder traversal of a binary tree without using recursion.
 */
int* inorderTraversal(struct TreeNode* root, int* returnSize) {
    // Initialize return size to 0
    *returnSize = 0;
    
    // Allocate memory for the result array
    int* res = malloc(sizeof(int) * 501);
    
    // Allocate memory for the stack to keep track of nodes
    struct TreeNode** stk = malloc(sizeof(struct TreeNode*) * 501);
    
    // Initialize top of the stack
    // variable top to keep track of the top of the stack. 
    int top = 0;
    
    // Iterative inorder traversal using a stack
    // The while loop continues until the current node  root  is NULL and the stack is empty (indicated by  top > 0 ).
    while (root != NULL || top > 0) {
        // Traverse left subtree and push nodes onto the stack 
        // a nested while loop to traverse the left subtree of the current node and pushes each node onto the stack. 
        while (root != NULL) {
            stk[top++] = root;
            root = root->left;
        }
        // Check if the stack is not empty before popping
        if (top > 0)
        {
        	// Once the left subtree is fully traversed
        	// Pop a node from the stack
        	root = stk[--top];
        
        	// Add the value of the popped node to the result array
        	res[(*returnSize)++] = root->val;
        
        	// Move to the right child of the popped node
        	root = root->right;
        }
    }
    // Free the memory allocated for the stack
    free(stk);
    
    // Return the result array containing inorder traversal
    return res;
}

复杂度分析

时间复杂度：O(n)，其中 n 为二叉树节点的个数。二叉树的遍历中每个节点会被访问一次且只会被访问一次。

空间复杂度：O(n)。空间复杂度取决于栈深度，而栈深度在二叉树为一条链的情况下会达到 O(n)的级别。

方法三：Morris 中序遍历

思路与算法

Morris 遍历算法是另一种遍历二叉树的方法，它能将非递归的中序遍历空间复杂度降为 O(1)。

Morris 遍历算法整体步骤如下（假设当前遍历到的节点为 xxx）：

如果 xxx 无左孩子，先将 xxx 的值加入答案数组，再访问 xxx 的右孩子，即 x=x.right。
如果 xxx 有左孩子，则找到 xxx 左子树上最右的节点（即左子树中序遍历的最后一个节点，xxx 在中序遍历中的前驱节点），我们记为 predecessor。根据 predecessor 的右孩子是否为空，进行如下操作。
- 如果 predecessor 的右孩子为空，则将其右孩子指向 xxx，然后访问 xxx 的左孩子，即 x=x.left。
- 如果 predecessor\ 的右孩子不为空，则此时其右孩子指向 xxx，说明我们已经遍历完 xxx 的左子树，我们将 predecessor 的右孩子置空，将 xxx 的值加入答案数组，然后访问 xxx 的右孩子，即 x=x.right。
重复上述操作，直至访问完整棵树。

其实整个过程我们就多做一步：假设当前遍历到的节点为 x，将 x 的左子树中最右边的节点的右孩子指向 x，这样在左子树遍历完成后我们通过这个指向走回了 x，且能通过这个指向知晓我们已经遍历完成了左子树，而不用再通过栈来维护，省去了栈的空间复杂度。

代码

/**
 * Definition for a binary tree node.
 */
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};
/**
 * The algorithm uses a predecessor node to establish temporary links 
 * between nodes to simulate the recursive call stack 
 * that would be used in a recursive inorder traversal. 
 * This approach allows for an iterative inorder traversal of the binary tree.
 */
int* inorderTraversal(struct TreeNode* root, int* returnSize) {
    // Allocate memory for the result array
    int* res = malloc(sizeof(int) * 501);
    
    // Initialize return size to 0
    *returnSize = 0;
    
    // Initialize predecessor node to NULL
    struct TreeNode* predecessor = NULL;
    
    // Traverse the tree in inorder without using recursion
    while (root != NULL) {
        // If the current node has a left child
        if (root->left != NULL) {
            // Find the predecessor node, which is the rightmost node in the left subtree
            predecessor = root->left;
            while (predecessor->right != NULL && predecessor->right != root) {
                predecessor = predecessor->right;
            }
            
            // If predecessor's right child is NULL, establish a link and move to the left child
            if (predecessor->right == NULL) {
                predecessor->right = root;
                root = root->left;
            }
            // If the left subtree has been visited, disconnect the link and move to the right child
            else {
                res[(*returnSize)++] = root->val;
                predecessor->right = NULL;
                root = root->right;
            }
        }
        // If there is no left child, visit the current node and move to the right child
        else {
            res[(*returnSize)++] = root->val;
            root = root->right;
        }
    }
    
    // Return the result array containing inorder traversal
    return res;
}

复杂度分析

时间复杂度：O(n)，其中 n 为二叉树的节点个数。Morris 遍历中每个节点会被访问两次，因此总时间复杂度为 O(2n)=O(n)。

空间复杂度：O(1)。

作者：力扣官方题解
链接：https://leetcode.cn/problems/binary-tree-inorder-traversal/solutions/412886/er-cha-shu-de-zhong-xu-bian-li-by-leetcode-solutio/
来源：力扣（LeetCode）
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