LinCode 刷题 之二叉树最小深度

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

 1 public class Solution {
 2     /**
 3      * @param root: The root of binary tree.
 4      * @return: An integer.
 5      */
 6     public int minDepth(TreeNode root) {
 7        Queue<TreeNode> queue=new LinkedList<TreeNode>();
 8     
 9         int num;
10         if(root==null){
11             num=0;
12         }else{
13             num=1;
14             queue.add(root);
15         }
16         while(!queue.isEmpty()){
17             int size = queue.size();
18             //遍历本层级的所有元素
19             for(int i=0;i<size;i++){
20                 TreeNode node = queue.poll();
21                 if(node.left==null&&node.right==null){
22                     queue=new LinkedList<>();//找到最深节点返回
23                    return num;
24                 }else{
25                     if(node.left!=null){
26                         queue.offer(node.left);
27                     }
28                     
29                     if(node.right!=null){
30                         queue.offer(node.right);
31                     }
32                 }
33             }
34             //层级+1
35             num++;
36             
37         }
38         
39         return num;
40     }
41 }