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 0;//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;
    }
};

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

　　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