111. 二叉树的最小深度 简单 广搜 递归

  1. 二叉树的最小深度
    给定一个二叉树,找出其最小深度。

最小深度是从根节点到最近叶子节点的最短路径上的节点数量。

说明:叶子节点是指没有子节点的节点。

示例 1:

输入:root = [3,9,20,null,null,15,7]
输出:2
示例 2:

输入:root = [2,null,3,null,4,null,5,null,6]
输出:5

提示:

树中节点数的范围在 [0, 105] 内
-1000 <= Node.val <= 1000
通过次数392,013提交次数778,508

广搜
import java.util.LinkedList;

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int minDepth(TreeNode root) {
        if(root==null)return 0;
        Queue<TreeNode> q=new LinkedList<>();
        Queue<Integer> nq=new LinkedList<>();
        q.offer(root);
        nq.offer(1);
        while(!q.isEmpty()){
        	TreeNode t=q.poll();
        	int d=nq.poll();
        	if(t.left==null&&t.right==null)return d;
        	if(t.left!=null){
        		q.offer(t.left);
        		nq.offer(d+1);
        	}
        	if(t.right!=null){
        		q.offer(t.right);
        		nq.offer(d+1);
        	}
        }
        return 0;
    }
}
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int minDepth(TreeNode root) {
        if(root==null)return 0;
        int l=minDepth(root.left);
        int r=minDepth(root.right);
        if(l==0&&r==0)return 1;
        if(l==0||r==0)return Math.max(l,r)+1;
        return Math.min(l,r)+1;
    }
}
posted @ 2022-11-17 23:01  林动  阅读(12)  评论(0编辑  收藏  举报