Leetcode—111.二叉树的最小深度【简单】

编程入门行业动态更新时间:2024-10-28 10:24:32

Leetcode—111.二叉树的<a href=https://www.elefans.com/category/jswz/34/1769671.html style= 最小深度【简单】"/>

Leetcode—111.二叉树的最小深度【简单】

2023每日刷题（十八）

Leetcode—111.二叉树的最小深度

DFS实现代码

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     struct TreeNode *left;*     struct TreeNode *right;* };*/
int minDepth(struct TreeNode* root) {if(root == NULL) {return 0;}if(root->left == NULL) {return minDepth(root->right) + 1;}if(root->right == NULL) {return minDepth(root->left) + 1;}   int left = minDepth(root->left);int right = minDepth(root->right);return(left < right ? left: right) + 1;
}

运行结果

BFS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:int minDepth(TreeNode* root) {if(root == NULL) {return 0;}int ans = 0;queue<TreeNode*> qu;qu.push(root);while(!qu.empty()) {ans++;int n = qu.size();for(int i = 0; i < n; i++) {TreeNode* p = qu.front();qu.pop();// 遇到第一个叶子结点返回ansif(p->left == NULL && p->right == NULL) {return ans;} if(p->left != NULL) {qu.push(p->left);}if(p->right != NULL) {qu.push(p->right);}}}return 0;}
};