题目描述

思路

利用二叉树的先序遍历，每次递归遍历时将当前节点的左右子节点交换即可

复杂度

时间复杂度:

$O(n)$；其中$n$为树节点的个数

空间复杂度:

$O(heigh)$；其中$height$为树的高度

Code

class Solution {
public:/**** @param root The root of binary tree* @return TreeNode**/TreeNode *invertTree(TreeNode *root) {traverse(root);return root;}/*** Binary tree traversal function** @param root The root of binary tree*/void traverse(TreeNode *root) {if (root == nullptr) {return;}//All each node needs to do is swap its left and right child nodesTreeNode *tmp = root->left;root->left = root->right;root->right = tmp;//Traverse the nodes of the left and right subtreetraverse(root->left);traverse(root->right);}
};