问题描述:
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,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
原问题链接:https://leetcode.com/problems/binary-tree-level-order-traversal-ii/
问题分析
这个问题和前面树的层次化遍历的问题基本上一样,唯一有一点差别的地方就是在遍历完之后将结果倒置一下就可以了。
详细的代码实现如下:
/** * 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>> result = new ArrayList<>(); if(root == null) return result; List<TreeNode> list = new ArrayList<>(); list.add(root); while(!list.isEmpty()) { List<Integer> intList = new ArrayList<>(); List<TreeNode> temp = new ArrayList<>(); for(TreeNode node : list) { intList.add(node.val); if(node.left != null) temp.add(node.left); if(node.right != null) temp.add(node.right); } result.add(intList); list = temp; } Collections.reverse(result); return result; } }