一、题目
LeetCode-226. 翻转二叉树
链接：https://leetcode-cn.com/problems/invert-binary-tree/
难度：简单
翻转一棵二叉树。
示例：
输入：
4
/ \
2 7
/ \ / \
1 3 6 9
输出：
4
/ \
7 2
/ \ / \
9 6 3 1
二、解题思路
从根节点开始，递归地对树进行遍历。
交换每个节点的交换两棵子树的位置，即可完成以root 为根节点的整棵子树的翻转。
三、实现过程
c++
/**
* 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 nullptr;
}
TreeNode* left = invertTree(root->left);
TreeNode* right = invertTree(root->right);
root->left = right;
root->right = left;
return root;
}
};
PHP
/**
* Definition for a binary tree node.
* class TreeNode {
* public $val = null;
* public $left = null;
* public $right = null;
* function __construct($val = 0, $left = null, $right = null) {
* $this->val = $val;
* $this->left = $left;
* $this->right = $right;
* }
* }
*/
class Solution {
/**
* @param TreeNode $root
* @return TreeNode
*/
function invertTree($root) {
if($root == null){
return null;
}
$left = $this->invertTree($root->left);
$right = $this->invertTree($root->right);
$root->left = $right;
$root->right = $left;
return $root;
}
}
JavaScript
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {TreeNode}
*/
var invertTree = function(root) {
if (root == null) {
return null;
}
let left = invertTree(root.left);
let right = invertTree(root.right);
root.left = right;
root.right = left;
return root;
};
四、小结
- 时间复杂度：O(N)，其中 NN 为二叉树节点的数目。
- 空间复杂度：O(1)
