GuilinDev

Lc0111

05 August 2008

111 Minimum Depth of Binary Tree

原题概述

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Note: A leaf is a node with no children.

Example:

Given binary tree ‘[3,9,20,null,null,15,7]’,

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    3
   / \
  9  20
    /  \
   15   7

return its minimum depth = 2.

题意和分析

同样的递归解法，与104类似，需要检查下是否到达叶子结点。

代码

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/**
 * 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;
        }
        if (root.left == null && root.right == null) {
            return 1;
        }
        
        // 判断左右子树是否为null，防止最后在Math.min的时候返回最已经为null的左子树或右子树(较小值为0)
        if (root.left == null) {
            return minDepth(root.right) + 1;
        }
        if (root.right == null) {
            return minDepth(root.left) + 1;
        }
        return Math.min(minDepth(root.left), minDepth(root.right)) + 1;
    }
}

BFS迭代解法锻炼思维

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/**
 * 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 depth = 1;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        
        while (!queue.isEmpty()) {
            int currentSize = queue.size();
            
            for (int i = 0; i < currentSize; i++) {
                TreeNode node = queue.poll();
                if (node.left == null && node.right == null) {
                    return depth; // 遇到第一个叶子节点立即返回，确保最小深度
                }
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            depth++;
        }
        return depth;
    }
}
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