LeetCode226. Invert Binary Tree

文章目录

    • 一、题目
    • 二、题解
      • 2.1 前序遍历版本
      • 2.2 中序遍历版本
      • 2.3 后序遍历版本

一、题目

Given the root of a binary tree, invert the tree, and return its root.

Example 1:

Input: root = [4,2,7,1,3,6,9]
Output: [4,7,2,9,6,3,1]
Example 2:

Input: root = [2,1,3]
Output: [2,3,1]
Example 3:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100

二、题解

2.1 前序遍历版本

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(root == nullptr) return root;
        swap(root->left,root->right);
        invertTree(root->left);
        invertTree(root->right);
        return root;
    }
};

2.2 中序遍历版本

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(root == nullptr) return root;
        invertTree(root->left);
        swap(root->left,root->right);
        invertTree(root->left);
        return root;
    }
};

2.3 后序遍历版本

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(root == nullptr) return root;
        invertTree(root->left);
        invertTree(root->right);
        swap(root->left,root->right);
        return root;
    }
};

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