题目描述

请实现一个函数，用来判断一棵二叉树是不是对称的。

如果一棵二叉树和它的镜像一样，那么它是对称的。

样例

如下图所示二叉树[1,2,2,3,4,4,3,null,null,null,null,null,null,null,null]为对称二叉树：
    1
   / \
  2   2
 / \ / \
3  4 4  3

如下图所示二叉树[1,2,2,null,4,4,3,null,null,null,null,null,null]不是对称二叉树：
    1
   / \
  2   2
   \ / \
   4 4  3

算法

(二叉树，递归) $O(n)$

递归判断两个子树是否互为镜像。

两个子树互为镜像当且仅当：

1. 两个子树的根节点值相等；
2. 第一棵子树的左子树和第二棵子树的右子树互为镜像，且第一棵子树的右子树和第二棵子树的左子树互为镜像；

时间复杂度

从上到下每个节点仅被遍历一遍，所以时间复杂度是 $O(n)$。

参考文献

yxc大神的题

C++ 代码

/**
 * 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) {
        return root==NULL||dfs(root->left,root->right);
    }
    bool dfs(TreeNode*p,TreeNode*q){
        if(p==NULL||q==NULL)return p==NULL&&q==NULL;
        return p->val==q->val&&dfs(p->left,q->right)&&dfs(p->right,q->left);
    }
};

C 代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
bool dfs(struct TreeNode*p,struct TreeNode*q){
    if(p==NULL||q==NULL)return p==NULL&&q==NULL;
    return p->val==q->val&&dfs(p->left,q->right)&&dfs(p->right,q->left);
}
bool isSymmetric(struct TreeNode* root) {
    return root==NULL||dfs(root->left,root->right);
}

Java 代码

/**
 * 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) {
        return root==null||dfs(root.left,root.right);
    }
    public boolean dfs(TreeNode p,TreeNode q){
        if(p==null||q==null)return p==null&&q==null;
        return p.val==q.val&&dfs(p.left,q.right)&&dfs(p.right,q.left);
    }
}

Python 代码

# 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 isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        return root==None or self.dfs(root.left,root.right)
    def dfs(self,p,q):
        if p==None or q==None:
            return p==None and q==None
        return p.val==q.val and self.dfs(p.left,q.right)and self.dfs(p.right,q.left)

Javascript 代码

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var dfs=function(p,q){
    if(p==null||q==null)return p==null&&q==null;
    return p.val==q.val&&dfs(p.left,q.right)&&dfs(p.right,q.left);
};
var isSymmetric = function(root) {
    return root==null||dfs(root.left,root.right)
};

Python3 代码

# 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 isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        return root==None or self.dfs(root.left,root.right)
    def dfs(self,p,q):
        if(p==None or q==None):
            return p==None and q==None
        return p.val==q.val and self.dfs(p.left,q.right)and self.dfs(p.right,q.left)

Go 代码

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func dfs(p*TreeNode,q*TreeNode)bool{
    if p==nil||q==nil{
        return p==nil&&q==nil
    }
    return p.Val==q.Val&&dfs(p.Left,q.Right)&&dfs(p.Right,q.Left)
}
func isSymmetric(root *TreeNode) bool {
    return root==nil||dfs(root.Left,root.Right)
}