Leetcode-102(Java) Binary Tree Level Order Traversal

標籤：

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]]

連結地址：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;        //建立一個隊列，用於存放所有節點        Queue<TreeNode> currentLevel = new LinkedList<TreeNode>();        currentLevel.add(root);        while(!currentLevel.isEmpty())        {            //建立一個List記錄當前層所有節點值            List<Integer> currentList = new LinkedList<Integer>();            int size = currentLevel.size();            for(int i = 0; i < size; i++)            {                //poll()擷取並移除此列表的頭結點                TreeNode currentNode = currentLevel.poll();                currentList.add(currentNode.val);                //記錄每一層的節點值到List<Integer>中                if(currentNode.left != null)                    currentLevel.add(currentNode.left);                if(currentNode.right != null)                    currentLevel.add(currentNode.right);                //將當前List<Integer>添加到List<List<Integer>>中            }            list.add(currentList);        }        return list;    }}

 

Leetcode-102(Java) Binary Tree Level Order Traversal