首页 > 代码库 > leetcode 107 Binary Tree Level Order Traversal II ----- java
leetcode 107 Binary Tree Level Order Traversal II ----- java
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,null,null,15,7]
,
3 / 9 20 / 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
与之前的题做法一样,就是插入的时候,每次在第一位插入。
/** * 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 = new ArrayList<List<Integer>>(); if( root == null) return list; Queue<TreeNode> tree = new LinkedList<TreeNode>(); tree.add(root); while( !tree.isEmpty() ){ List ans = new ArrayList<Integer>(); int size = tree.size(); for( int i = 0;i<size;i++){ TreeNode node = tree.poll(); ans.add(node.val); if( node.left != null ) tree.add(node.left); if( node.right != null) tree.add(node.right); } if( !ans.isEmpty() ) list.add(0,ans); } return list; } }
leetcode 107 Binary Tree Level Order Traversal II ----- java
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。