/**
<p>給定一個二叉樹，找出其最小深度。</p>
<p>最小深度是從根節點到最近葉子節點的最短路徑上的節點數量。</p>
<p><strong>說明：</strong>葉子節點是指沒有子節點的節點。</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>輸入：</strong>root = [3,9,20,null,null,15,7]
<strong>輸出：</strong>2
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>輸入：</strong>root = [2,null,3,null,4,null,5,null,6]
<strong>輸出：</strong>5
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li>樹中節點數的範圍在 <code>[0, 10<sup>5</sup>]</code> 內</li>
<li><code>-1000 <= Node.val <= 1000</code></li>
</ul>
<div><div>Related Topics</div><div><li>樹</li><li>深度優先搜索</li><li>廣度優先搜索</li><li>二叉樹</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)
leetcodebfs-111
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