[LeetCode] Invert Binary Tree

4
/   \
2     7
/ \   / \
1   3 6   9

4
/   \
7     2
/ \   / \
9   6 3   1

解题思路

实现代码

// 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;
}
};

// 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;
}
}

# 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