leetcode 101. 对称二叉树 Symmetric Tree（使用c++/java/python）

https://leetcode-cn.com/problems/symmetric-tree/

https://leetcode.com/problems/symmetric-tree/

递归。利用一个函数判断两颗树是否镜像对称，即需要满足：

1. 两树根节点相同
2. 树1的左子树与树2的右子树镜像
3. 树1的右子树与树2的左子树镜像

执行用时： c++ 8ms； java 15ms； python 48ms

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:
    bool isSymmetric(TreeNode* root) {
        if (root==NULL)
            return true;
        else
            return ismirror(root->left, root->right);
    }
    bool ismirror(TreeNode* n1,TreeNode* n2)
    {
        if (n1==NULL && n2==NULL)
            return true;
        if (n1==NULL || n2==NULL)
            return false;
        return (n1->val==n2->val)&&(ismirror(n1->left,n2->right))&&(ismirror(n1->right,n2->left));
    }
};

java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root==null)
            return true;
        else
            return ismirror(root.left,root.right);
    }
    public boolean ismirror(TreeNode root1,TreeNode root2)
    {
        if(root1==null && root2==null)
            return true;
        if(root1==null || root2==null)
            return false;
        return root1.val==root2.val && ismirror(root1.left,root2.right) && ismirror(root1.right,root2.left);
    }
}

python

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

class Solution:
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
        else:
            return self.ismirror(root.left,root.right)
        
    def ismirror(self,root1,root2):
        if root1 is None and root2 is None:
            return True
        if root1 is None or root2 is None:
            return False
        return root1.val==root2.val and self.ismirror(root1.left,root2.right) and self.ismirror(root1.right,root2.left)