leetcode minimum-depth-of-binary-tree

题目描述

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.

 

我的解题思路：问题的关键是遍历，那么就有深度优先和宽度优先的方法，深度优先使用递归，代码更加优雅，需要多多学习，而宽度优先需要队列操作，记住Queue是一个接口，LinkListed是Queue的一个实现类。

 

在下的代码 时间384ms 空间15100k

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.LinkedList;

public class Solution {
    public int run(TreeNode root) {
        if (root == null)
            return 0;

        LinkedList<TreeNode> a = new LinkedList<TreeNode>();
        root.val = 1;

        a.add(root);
        while (!a.isEmpty()){
            TreeNode node = a.poll();
            if (node.left != null){
                node.left.val = node.val + 1;
                a.add(node.left);
            }
            if (node.right != null){
                node.right.val = node.val + 1;
                a.add(node.right); 
            }
            if (node.left == null && node.right == null){
				return node.val;
            }
        }
        return 0;
    }
}

　　

类似的 时间267ms 空间14928k

import java.util.LinkedList;
import java.util.Queue;
public class Solution {
    public int run(TreeNode root) {
        if(root == null)
            return 0;
        if(root.left == null && root.right == null)
            return 1;
         
        int depth = 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()){
            int len = queue.size();
            depth++;
            for(int i = 0; i < len; i++){
                TreeNode cur = queue.poll();            
                if(cur.left == null && cur.right == null)
                    return depth;
                if(cur.left != null)
                    queue.offer(cur.left);
                if(cur.right != null)
                    queue.offer(cur.right);
            }               
        }
        return 0;
    }
}

　　

大神代码 时间232ms 空间14404k

public class Solution {
    public int run(TreeNode root) {
        if(root==null)
            return 0;
        if(root.left==null&&root.right==null)
            return 1;
        if(root.left==null)
            return run(root.right)+1;
        if(root.right==null)
            return run(root.left)+1;
        return Math.min(run(root.left),run(root.right))+1;
    }
}