Leetcode OJ 102 Binary Tree Level Order Traversal [Medium]

Leetcode OJ 102 Binary Tree Level Order Traversal [Medium]

《剑指offer》第4章-面试题32

题目描述：

Given a binary tree, return the level ordertraversal 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]

]

题目理解：

一层一层打印二叉树。

测试用例：

功能测试：完全二叉树；普通二叉树；只有左子树的二叉树；只有右子树的二叉树；

特殊输入：二叉树根节点是null

分析：

1.  ；

 

错误解答：

   /** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */class Solution {    public List<List<Integer>> levelOrder(TreeNode root) {        List<List<Integer>> result = new ArrayList<List<Integer>>();        List<TreeNode> buffer = new ArrayList<TreeNode>();        TreeNode current = root;        List<Integer> subList = new ArrayList<Integer>();        int count=0,countNext=1;        int levelNum = 0;                while(current!= null || !buffer.isEmpty()){            if(current!=null){                subList.add(current.val);                levelNum += 1;                buffer.add(current.left);                buffer.add(current.right);            }            else{levelNum += 2;}            if(levelNum == countNext){                result.add(new ArrayList<Integer>(subList));                subList.clear();                levelNum = 0;                countNext *= 2;            }            //get next node;            if(!buffer.isEmpty()){                current = buffer.get(0);                buffer.remove(0);            }            else{                current = null;            }        }        return result;    }}