LeetCode 102. Binary Tree Level Order Traversal

原题链接在这里：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]
]

题解：

是一道BFS，利用queue，先进先出，并利用curCount来判断是否走完了当前一层。当curCount = 0 时表示当前一层走完，要把当前list加入res里，再刷新list，问题是不要忘记同时刷新curCount和nextCount.

Time Complexity: O(n). Space: O(n).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if(root == null){
            return res;
        }
        
        LinkedList<TreeNode> que = new LinkedList<TreeNode>();
        que.add(root);
        int curCount = 1;
        int nextCount = 0;
        List<Integer> item = new ArrayList<Integer>();
        while(!que.isEmpty()){
            TreeNode tn = que.poll();
            item.add(tn.val);
            curCount--;
            
            if(tn.left != null){
                que.add(tn.left);
                nextCount++;
            }
            if(tn.right != null){
                que.add(tn.right);
                nextCount++;
            }
            
            if(curCount == 0){
                res.add(new ArrayList<Integer>(item));
                item = new ArrayList<Integer>();
                curCount = nextCount;
                nextCount = 0;
            }
        }
        return res;
    }
}

 类似的有Check Completeness of a Binary Tree.