Leetcode—101.对称二叉树【简单】

2023每日刷题（十九）

Leetcode—101.对称二叉树

利用Leetcode101.对称二叉树的思想的实现代码

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

运行结果

层次遍历实现代码

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:bool isSymmetric(TreeNode* root) {if(root == NULL) {return true;}queue<TreeNode*> qu1, qu2;TreeNode* p1, *p2;qu1.push(root->left);qu2.push(root->right);while(!qu1.empty() && !qu2.empty()) {p1 = qu1.front();qu1.pop();p2 = qu2.front();qu2.pop();if((p1 && !p2) || (!p1 && p2)) {return false;}if(p1 != NULL && p2 != NULL) {if(p1->val != p2->val) {return false;}qu1.push(p1->left);qu1.push(p1->right);qu2.push(p2->right);qu2.push(p2->left);}}return true;}
};

运行结果

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