标签:bool sem 12px public clear 节点 1.4 article border
题目:
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]
]
题意:
给定一棵二叉树,依照层顺序遍历二叉树全部的节点(即 从左向右 一层层地)
比方,给定二叉树{3,9,20,#,#,15,7}
,
3 / 9 20 / 15 7返回它的层遍历为:
[
[3],
[9,20],
[15,7]
]
算法分析:
该题是对二叉树进行层次优先遍历,层次遍历主要採用队列的形式进行存储,通过将每一个节点的左孩子和右孩子放入队列中,然后每次从队列中取出元素就可以。比較好理解,直接上代码了。
AC代码:
public class Solution { private static TreeNode root; public static ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) { ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>(); if (root == null) { return res; } ArrayList<Integer> tmp = new ArrayList<Integer>(); Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.offer(root); int num; boolean reverse = false; while (!queue.isEmpty()) { num = queue.size(); //每次通过这个确定终于的出队数目 tmp.clear(); for (int i = 0; i < num; i++) //队列中出1个父。进两个子;出2个父,进4个子;出4个父。进8个子 { TreeNode node = queue.poll(); tmp.add(node.val); if (node.left != null) queue.offer(node.left); if (node.right != null) queue.offer(node.right); } res.add(new ArrayList<Integer>(tmp)); } return res; } }
[LeetCode][Java] Binary Tree Level Order Traversal
标签:bool sem 12px public clear 节点 1.4 article border
原文地址:http://www.cnblogs.com/clnchanpin/p/7043944.html