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invert-binary-tree

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Invert Binary Tree – LeetCode 226

Problem

Description

Invert a binary tree.

Answer

Original

Code

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

思路

看题说话。时间复杂度$O(n)$，空间复杂度$O(n)$。
耗时$3$ ms，排名$39.58\%$

Better

思路

简单的写成BFS，时间复杂度$O(n)$，空间复杂度$O(n)$。
耗时$3$ ms，排名$39.58\%$

Code

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if (!root) return NULL;
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            TreeNode *node = q.front(); q.pop();
            TreeNode *tmp = node->left;
            node->left = node->right;
            node->right = tmp;
            if (node->left) q.push(node->left);
            if (node->right) q.push(node->right);
        }
        return root;
    }
};