Symmetric Tree

问题:判断二叉树是否为镜像二叉树
分析:递归判断,根节点单独判断,然后递归左结点和右结点,之后每次一起递归左结点的左结点和右结点的右结点比较,左结点的右结点和右结点的左结点比较

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool dfs(TreeNode *left,TreeNode *right)
    {
        if(!left && !right) return true;//需要先判断这个,不然val那个可能会出现RE
        if((left && !right) || (!left && right) || (left->val!=right->val)) return false;
        return dfs(left->left,right->right) && dfs(left->right,right->left);
    }
    bool isSymmetric(TreeNode *root) {
        if(root==NULL || (!root->left && !root->right)) return true;
        return dfs(root->left,root->right);
    }
};

  

原文地址:https://www.cnblogs.com/zsboy/p/3890454.html