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;
    }
}