分析

难度 中

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来源

https://leetcode.com/problems/binary-tree-level-order-traversal/
 

题目

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3

   / \

  9  20

    /  \

   15   7

return its level order traversal as:

[

  [3],

  [9,20],

  [15,7]

]

解答

 1 package LeetCode;
 2 
 3 import java.util.*;
 4 
 5 public class L102_BinaryTreeLevelOrderTraversal {
 6     public TreeNode makeBinaryTreeByArray(Integer[] array,int index){
 7         if(index<array.length&&array[index]!=null){
 8             int value=array[index];
 9             TreeNode t=new TreeNode(value);
10             array[index]=0;
11             t.left=makeBinaryTreeByArray(array,index*2+1);
12             t.right=makeBinaryTreeByArray(array,index*2+2);
13             return t;
14         }else
15             return null;
16     }
17 
18     public List<List<Integer>> levelOrder(TreeNode root) {
19         List<Integer> levelList=new ArrayList<Integer>();
20         List<List<Integer>> result=new ArrayList<List<Integer>>();
21         if(root==null){
22             return result;
23         }
24         Queue<TreeNode> queue=new LinkedList<TreeNode>();
25         queue.offer(root);
26         int curCount=1;//记录当前层节点数
27         int nextCount=0;//记录下一层节点数
28         int outCount=0;//记录出队列节点数
29         while(!queue.isEmpty()){
30             TreeNode node=queue.poll();
31             levelList.add(node.val);
32             outCount++;
33             if(node.left!=null){
34                 queue.offer(node.left);
35                 nextCount++;
36             }
37             if(node.right!=null){
38                 queue.offer(node.right);
39                 nextCount++;
40             }
41             if(outCount==curCount)//当前层全部出队列
42             {
43                 if(!levelList.isEmpty())
44                     result.add(levelList);
45                 levelList=new ArrayList<Integer>();
46                 curCount=nextCount;
47                 nextCount=0;
48                 outCount=0;
49             }
50         }
51         return result;
52     }
53 
54     public static void main(String[] args){
55         L102_BinaryTreeLevelOrderTraversal l102=new L102_BinaryTreeLevelOrderTraversal();
56         Integer[] values={1,2,3,4,5,6,7};
57         TreeNode root=l102.makeBinaryTreeByArray(values,0);
58         List<List<Integer>> res=l102.levelOrder(root);
59         System.out.println(res.toString());
60     }
61 }

 

 

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