Binary-tree-level-order-traversal

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/Neo233/article/details/81208675

题目描述

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]
]

confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.*;
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> a = new ArrayList<ArrayList<Integer>>();
        if(root == null){
            return a;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while(!queue.isEmpty()){
            ArrayList<Integer> level = new ArrayList<Integer>();
            int levelnum = queue.size();
            for(int i = 0;i < levelnum;i++){
                if(queue.peek().left != null){
                    queue.offer(queue.peek().left);
                }
                if(queue.peek().right != null){
                    queue.offer(queue.peek().right);
                }
                level.add(queue.poll().val);
            }
            a.add(level);
        }
        return a;
    }
}