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leetcodebfs-111

西特张 2022-07-12 阅读 54
/**
<p>给定一个二叉树,找出其最小深度。</p>

<p>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。</p>

<p><strong>说明:</strong>叶子节点是指没有子节点的节点。</p>

<p> </p>

<p><strong>示例 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/12/ex_depth.jpg" style="width: 432px; height: 302px;" />
<pre>
<strong>输入:</strong>root = [3,9,20,null,null,15,7]
<strong>输出:</strong>2
</pre>

<p><strong>示例 2:</strong></p>

<pre>
<strong>输入:</strong>root = [2,null,3,null,4,null,5,null,6]
<strong>输出:</strong>5
</pre>

<p> </p>

<p><strong>提示:</strong></p>

<ul>
<li>树中节点数的范围在 <code>[0, 10<sup>5</sup>]</code> 内</li>
<li><code>-1000 <= Node.val <= 1000</code></li>
</ul>
<div><div>Related Topics</div><div><li>树</li><li>深度优先搜索</li><li>广度优先搜索</li><li>二叉树</li></div></div><br><div><li>👍 784</li><li>👎 0</li></div>
*/

//leetcode submit region begin(Prohibit modification and deletion)


import java.util.LinkedList;
import java.util.Queue;

/**
* 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 {
//普通的bfs
public int minDepth(TreeNode root) {
if (root == null) {
return 0;
}

Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);

int depth = 1;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode poll = queue.poll();
if(poll.left==null && poll.right==null){
return depth;
}

if (poll.left != null) {
queue.offer(poll.left);
}
if (poll.right != null) {
queue.offer(poll.right);
}
}
depth++;
}
return depth;
}
}
//leetcode submit region end(Prohibit modification and deletion)

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