leetcode 111. Minimum Depth of Binary Tree

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Note: A leaf is a node with no children.

 

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 2

Example 2:

Input: root = [2,null,3,null,4,null,5,null,6]
Output: 5

 

Constraints:

  • The number of nodes in the tree is in the range [0, 105].
  • -1000 <= Node.val <= 1000

题目大意：求二叉树的最小深度。

难度：简单题

方法一：深度优先搜索（DFS）

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 8  *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 9  *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
10  * };
11  */
12 class Solution {
13 public:
14     int minDepth(TreeNode* root) {
15         if (root == nullptr) return 0;
16         int left = minDepth(root->left);
17         int right = minDepth(root->right);
18         return (left == 0 || right == 0) ? left + right + 1 : min(left, right) + 1;
19     }
20 };

 

方法二：广度优先搜索（BFS）

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 8  *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 9  *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
10  * };
11  */
12 class Solution {
13 public:
14     int minDepth(TreeNode *root) {
15         queue<TreeNode*> q;
16         TreeNode *temp;
17         if (root == nullptr) return 0;
18         q.push(root);
19         bool flag = false;
20         int d = 1;
21         while (!q.empty()) {
22             int len = q.size();
23             for (int i = 0; i < len; ++i) {
24                 temp = q.front();
25                 q.pop();
26                 if ((temp->left == nullptr) && (temp->right == nullptr)) {
27                     flag = true;
28                     break;
29                 }
30                 if (temp->left) q.push(temp->left);
31                 if (temp->right) q.push(temp->right);
32             }
33             if (flag) {
34                 break;
35             }
36             ++d;
37         }
38         return d;
39     }
40 };

 

posted @ 2020-10-22 22:37  琴影  阅读(87)  评论(0)  编辑  收藏  举报