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/*

Author: Andy, nkuwjg@gmail.com

Date: Dec 12, 2014

Problem: Binary Tree Level Order Traversal II

Difficulty: easy

Source: https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/

Notes:

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],

]

Solution: Queue version. On the basis of 'Binary Tree Level Order Traversal', reverse the final vector.

*/

/**

* Definition for binary tree

* 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>> res = new ArrayList<List<Integer>>();

if (root == null) return res;

Queue<TreeNode> q = new LinkedList<TreeNode>();

q.offer(root);

q.offer(null);

List<Integer> level = new ArrayList<Integer>();

while(true) {

TreeNode node = q.poll();

if (node != null) {

level.add(node.val);

if(node.left!=null) q.offer(node.left);

if(node.right!=null) q.offer(node.right);

} else {

res.add(level);

level = new ArrayList<Integer>();

if(q.isEmpty()==true) break;

q.offer(null);

}

}

Collections.reverse(res);

return res;

}

}