Leetcode-102. Binary Tree Level Order Traversal

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前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

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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,null,null,15,7],

    3   / \  9  20    /  \   15   7

return its level order traversal as:

[  [3],  [9,20],  [15,7]]

这个用两个队列先后维护树的不同层。Your runtime beats 36.62% of java submissions.

/** * 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>> results = new ArrayList<List<Integer>>();        Queue<TreeNode> queue1 = new LinkedList<TreeNode>();Queue<TreeNode> queue2 = new LinkedList<TreeNode>();queue1.offer(root);while(queue1.peek() != null || queue2.peek() != null){List<Integer> result = null;if(queue1.peek() != null){    result = new ArrayList<Integer>();    while(queue1.peek() != null){    TreeNode node = queue1.poll();    if(node != null) result.add(node.val);    if(node.left !=null)queue2.offer(node.left);    if(node.right != null)queue2.offer(node.right);    }    results.add(result);}if(queue2.peek() != null){    result = new ArrayList<Integer>();    while(queue2.peek() != null){    TreeNode node = queue2.poll();    if(node != null) result.add(node.val);    if(node.left != null)queue1.offer(node.left);    if(node.right != null)queue1.offer(node.right);    }    results.add(result);}}return results;    }}

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