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 nodes
        TreeNode *tmp = root->left;
        root->left = root->right;
        root->right = tmp;

        //Traverse the nodes of the left and right subtree
        traverse(root->left);
        traverse(root->right);
    }
};