LeetCode | 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]]

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */

 //层序遍历，使用队列来实现: //队列的前半部分是当前层节点，后半部分是下一层节点 //remove队头遍历其val，然后将其left与right(如果有的话)入队 //使用标记位nextLevelStarter来标记下一层的首个节点，以免在队列中遍历当前层时越界public class Solution {    public List<List<Integer>> levelOrder(TreeNode root) {        List<List<Integer>> result = new ArrayList<List<Integer>>();        if(root==null) return result;                Queue<TreeNode> nodeQueue = new LinkedList<TreeNode>();//注意Java中Queue是接口，不能直接实例化，而LinkedList实现了Queue接口        nodeQueue.add(root);        TreeNode nextLevelStarter;   //标记下一层的起始节点，以免在Queue中遍历是越界                while(!nodeQueue.isEmpty()){            List<Integer> thisLevelList = new ArrayList<Integer>();            nextLevelStarter = null;    //在开始遍历本层时，先把其置为null，然后找下一层的首节点赋予                        while(nodeQueue.isEmpty()==false && nodeQueue.peek()!=nextLevelStarter){//在Queue中遍历本层的节点                TreeNode curNode = nodeQueue.remove();   //peek()只返回头，remove()返回并删除头                thisLevelList.add(curNode.val);                if(curNode.left!=null) nodeQueue.add(curNode.left);                if(curNode.right!=null) nodeQueue.add(curNode.right);                                if(nextLevelStarter==null){     //说明还未找到下层的首节点，就进入寻找                    if(curNode.left!=null){                        nextLevelStarter = curNode.left;                    }else if(curNode.right!=null){                        nextLevelStarter = curNode.right;                    }                }            }            result.add(thisLevelList);        }                return result;    }}