leetcodebfs111

/**
<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=" 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>
*/

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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;
    }
}
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