111. 二叉树的最小深度 

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

 

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution 
11 {
12 public:
13     int minDepth(TreeNode* root) 
14     {
15         if (root == NULL) return 0;
16         queue<TreeNode*> q;
17         q.push(root);
18         // root 本身就是一层，depth 初始化为 1
19         int depth = 1;
20 
21         while (!q.empty()) 
22         {
23             int sz = q.size();
24             /* 将当前队列中的所有节点向四周扩散 */
25             for (int i = 0; i < sz; i++) 
26             {
27                 TreeNode* cur = q.front();
28                 q.pop();
29                 /* 判断是否到达终点 */
30                 if (cur->left == NULL && cur->right == NULL)  
31                     return depth;
32                 /* 将 cur 的相邻节点加入队列 */
33                 if (cur->left) q.push(cur->left);
34                 if (cur->right) q.push(cur->right);
35             }
36             /* 这里增加步数 */
37             depth++;
38         }
39         return depth;
40     }
41 };

 

posted @ 2020-04-01 14:52  Jinxiaobo0509  阅读(138)  评论(0)    收藏  举报