Leetcode-102(Java) Binary Tree Level Order Traversal

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]
]

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

 

/**
 * 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>> list = new LinkedList<List<Integer>>();
        if(root == null)
            return list;
        //创建一个队列，用于存放所有节点
        Queue<TreeNode> currentLevel = new LinkedList<TreeNode>();
        currentLevel.add(root);
        while(!currentLevel.isEmpty())
        {
            //创建一个List记录当前层所有节点值
            List<Integer> currentList = new LinkedList<Integer>();
            int size = currentLevel.size();
            for(int i = 0; i < size; i++)
            {
                //poll()获取并移除此列表的头结点
                TreeNode currentNode = currentLevel.poll();
                currentList.add(currentNode.val);
                //记录每一层的节点值到List<Integer>中
                if(currentNode.left != null)
                    currentLevel.add(currentNode.left);
                if(currentNode.right != null)
                    currentLevel.add(currentNode.right);
                //将当前List<Integer>添加到List<List<Integer>>中
            }
            list.add(currentList);
        }
        return list;
    }
}