LeetCode Binary Tree Level Order Traversal II
2015-05-05 09:52
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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
return its bottom-up level order traversal as:
confused what
read more on how binary tree is serialized on OJ.
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
The above binary tree is serialized as
题意:倒着层次遍历一棵树。
思路:还是一样,最后倒置一下就行了
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] ]
confused what
"{1,#,2,3}"means? >
read more on how binary tree is serialized on OJ.
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1 / \ 2 3 / 4 \ 5
The above binary tree is serialized as
"{1,2,3,#,#,4,#,#,5}".
题意:倒着层次遍历一棵树。
思路:还是一样,最后倒置一下就行了
/** * 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>> ans = new ArrayList<List<Integer>>(); if (root == null) return ans; Queue<TreeNode> q = new LinkedList<TreeNode>(); q.add(root); int count = 1, level = 0; while (!q.isEmpty()) { List<Integer> tmp = new ArrayList<Integer>(); level = 0; for (int i = 0; i < count; i++) { TreeNode t = q.poll(); tmp.add(t.val); if (t.left != null) { q.add(t.left); ++level; } if (t.right != null) { q.add(t.right); ++level; } } count = level; ans.add(tmp); } Collections.reverse(ans); return ans; } }
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