Binary Tree Level Order Traversal - LeetCode

本题最重要的是

for (int i = 0; i < length; i++)  

和

 for (int j = 0; j < length; j++)           {                       siblings.remove(0);           }  

保证了逐层的输出。

以下是AC代码：

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {         ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>> ();      ArrayList<TreeNode> siblings = new ArrayList<TreeNode> ();            if ( root != null)          siblings.add(root);            while ( !siblings.isEmpty() )      {           ArrayList<Integer> tmp = new ArrayList<Integer> ();           int length = siblings.size();                    for (int i = 0; i < length; i++)           {               tmp.add(siblings.get(i).val);                              if(siblings.get(i).left != null)                   siblings.add(siblings.get(i).left);                              if(siblings.get(i).right != null)                   siblings.add(siblings.get(i).right);                }                      for (int j = 0; j < length; j++)           {                       siblings.remove(0);           }                           results.add(tmp);               }        return results;    }}

这是层序输出的基本描述。

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,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its level order traversal as:

[  [3],  [9,20],  [15,7]]