2023每日刷题（十八）

Leetcode—100.相同的树

递归实现代码

/*** 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->left) && isSameTree(p->right, q->right)) {return true;}return false;
}

运行结果

先序序列化实现代码

/*** 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) {}* };*/#define nulll -1class Solution {
public:void preorderseq(TreeNode* b, vector<int> &pre) {if(b) {// 根pre.push_back(b->val);// 左preorderseq(b->left, pre);// 右preorderseq(b->right, pre);} else {pre.push_back(nulll);}}bool isSameTree(TreeNode* p, TreeNode* q) {if(p == nullptr && q == nullptr) {return true;}if((!p && q) || (p && !q)) {return false;}vector<int> pre1, pre2;preorderseq(p, pre1);preorderseq(q, pre2);if(pre1.size() != pre2.size()) {return false;}for(int i = 0; i < pre1.size(); i++) {if(pre1[i] != pre2[i]) {return false;}}return true;}
};

运行结果

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