Java [Leetcode 107]Binary Tree Level Order Traversal II
题目描述:
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree {3,9,20,#,#,15,7}
,
3 / \ 9 20 / \ 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
解题思路:
运用广度优先搜索方法,运用队列的方法,每次遍历一层的叶节点,并把下一层的叶节点加入到队列中。每次遍历一层节点结束时候,将该层节点组成的list放到整体的list之前,实现由底到高的排列。
代码如下:
/** * 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>> levelOrderBottom(TreeNode root) { List<List<Integer>> list = new LinkedList<List<Integer>>(); Queue<TreeNode> queue = new LinkedList<TreeNode>(); if(root == null) return list; queue.offer(root); while(!queue.isEmpty()){ int num = queue.size(); List<Integer> levelList = new LinkedList<Integer>(); for(int i = 0; i < num; i++){ if(queue.peek().left != null) queue.offer(queue.peek().left); if(queue.peek().right != null) queue.offer(queue.peek().right); levelList.add(queue.poll().val); } list.add(0, levelList); } return list; } }