力扣-剑指 Offer 28. 对称的二叉树（C++）- 递归+深度遍历

桃扇骨 2022-09-09 09:29 122阅读 0赞

题目链接：[https://leetcode-cn.com/problems/dui-cheng-de-er-cha-shu-lcof/][https_leetcode-cn.com_problems_dui-cheng-de-er-cha-shu-lcof]  
题目如下：  
![在这里插入图片描述][watermark_type_ZHJvaWRzYW5zZmFsbGJhY2s_shadow_50_text_Q1NETiBASkluZ2xlczEyMw_size_20_color_FFFFFF_t_70_g_se_x_16]  
![在这里插入图片描述][watermark_type_ZHJvaWRzYW5zZmFsbGJhY2s_shadow_50_text_Q1NETiBASkluZ2xlczEyMw_size_20_color_FFFFFF_t_70_g_se_x_16 1]

/**
     * 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 pre(TreeNode* left,TreeNode* right){
        
            if(left==NULL&&right==NULL) return true;//两个子节点均为空
    
            if(left==NULL&&right!=NULL) return false;//左右子节点一个为空，一个不为空
            if(left!=NULL&&right==NULL) return false;
            if(left->val!=right->val) return false;//左右节点的数值不相同
    
            return pre(left->left,right->right)&&pre(left->right,right->left);
            
        }
    
        bool isSymmetric(TreeNode* root) {
        
            if(root==NULL) return true;
    
            return pre(root->left,root->right);
        }
    };

[https_leetcode-cn.com_problems_dui-cheng-de-er-cha-shu-lcof]: https://leetcode-cn.com/problems/dui-cheng-de-er-cha-shu-lcof/
[watermark_type_ZHJvaWRzYW5zZmFsbGJhY2s_shadow_50_text_Q1NETiBASkluZ2xlczEyMw_size_20_color_FFFFFF_t_70_g_se_x_16]: /images/20220829/7ec9bda8108d424ea928d05c9ea79587.png
[watermark_type_ZHJvaWRzYW5zZmFsbGJhY2s_shadow_50_text_Q1NETiBASkluZ2xlczEyMw_size_20_color_FFFFFF_t_70_g_se_x_16 1]: /images/20220829/99e1e4d4880b4b63a412b17b81930815.png