LeetCode.111. 二叉树的最小深度

​​​​​​111. 二叉树的最小深度

难度：easy

/**
 * 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) {
        // int minDepth = Integer.MAX_VALUE;
        if (root == null) {
            return 0;
        }
        Queue queue = new LinkedList<>();
        queue.offer(root);
        int depth = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            // 层数+1
            depth++;
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                // 一个节点不存在左节点和右节点就可以计算深度；
                if (node.left == null && node.right == null) {                   
                    // minDepth = Math.min(minDepth, depth);
                    return depth;
                }
                if (node.left != null) {
                    queue.offer(node.left); 
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }             
            }
        }
 
        return minDepth;
    }
}

在leetcode看到一个不错的题解，通过自行封装QueueNode加入depth信息，随时可以获得最短的深度。

class Solution {
    class QueueNode {
        TreeNode node;
        int depth;
 
        public QueueNode(TreeNode node, int depth) {
            this.node = node;
            this.depth = depth;
        }
    }
 
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
 
        Queue queue = new LinkedList();
        queue.offer(new QueueNode(root, 1));
        while (!queue.isEmpty()) {
            QueueNode nodeDepth = queue.poll();
            TreeNode node = nodeDepth.node;
            int depth = nodeDepth.depth;
            if (node.left == null && node.right == null) {
                return depth;
            }
            if (node.left != null) {
                queue.offer(new QueueNode(node.left, depth + 1));
            }
            if (node.right != null) {
                queue.offer(new QueueNode(node.right, depth + 1));
            }
        }
 
        return 0;
    }
}