LeetCode Invert Binary Tree 反转二叉树

思路：递归解决，在返回root前保证该点的两个孩子已经互换了。注意可能给一个Null。

C++

 /**
  * 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 ;//重要在这而已
         if(root->left)    invertTree(root->left);
         if(root->right)   invertTree(root->right);
         TreeNode* tmp=root->right;
         root->right=root->left;
         root->left=tmp;
         return root;
     }
 };

AC代码

python3

　　递归

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

 class Solution(object):
     def invertTree(self, root):
         """
         :type root: TreeNode
         :rtype: TreeNode
         """
         if root!=None:
             root.left, root.right= self.invertTree(root.right), self.invertTree(root.left)
         return root

AC代码

　　迭代

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

 class Solution(object):
     def invertTree(self, root):
         """
         :type root: TreeNode
         :rtype: TreeNode
         """
         stack=[root]
         while stack!=[]:
             back=stack.pop()
             if back!=None:
                 back.left, back.right= back.right, back.left
                 stack.extend([back.left,back.right])#也可以写成stack+=back.left,back.right

         return root

AC代码