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[LeetCode] Invert Binary Tree

时间:2015-06-18 17:22:30      阅读:82      评论:0      收藏:0      [点我收藏+]

标签:leetcode

Invert a binary tree.

     4
   /     2     7
 / \   / 1   3 6   9

to

     4
   /     7     2
 / \   / 9   6 3   1

解题思路

递归

实现代码

C++:

// Runtime: 3 ms
/**
 * 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 != NULL)
        {
            TreeNode *temp = root->left;
            root->left = root->right;
            root->right = temp;
            invertTree(root->left);
            invertTree(root->right);
        }
        return root;
    }
};

Java:

// Runtime: 238 ms
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root != null){
            TreeNode temp = root.left;
            root.left = root.right;
            root.right = temp;
            invertTree(root.left);
            invertTree(root.right);
        }

        return root;
    }
}

Python:

# Runtime: 52 ms
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param {TreeNode} root
    # @return {TreeNode}
    def invertTree(self, root):
        if root != None:
            root.left, root.right = root.right, root.left
            self.invertTree(root.left)
            self.invertTree(root.right)
        return root

[LeetCode] Invert Binary Tree

标签:leetcode

原文地址:http://blog.csdn.net/foreverling/article/details/46548853

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