【leetcode】【107】Binary Tree Level Order Traversal II

一、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,#,#,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.

二、问题分析

这道题与【leetcode】【102】Binary Tree Level Order Traversal几乎一样，只不过因为是从后往前输出，所以可以利用stack.

三、Java AC代码

public List<List<Integer>> levelOrderBottom(TreeNode root) {
LinkedList<List<Integer>> list = new LinkedList<List<Integer>>();
if (root==null) {
return list;
}
LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
LinkedList<Integer> listItem = new LinkedList<Integer>();
queue.add(root);
int curLevelItemNums = 1;
int nextLevelItemNums = 0;
while(!queue.isEmpty()){
TreeNode node = queue.poll();
TreeNode leftNode = node.left;
TreeNode rightNode = node.right;
curLevelItemNums--;
listItem.add(node.val);
if (leftNode!=null) {
queue.add(leftNode);
nextLevelItemNums++;
}
if (rightNode!=null) {
queue.add(rightNode);
nextLevelItemNums++;
}
if (curLevelItemNums == 0) {
curLevelItemNums = nextLevelItemNums;
nextLevelItemNums = 0;
list.push(listItem);
listItem = new LinkedList<Integer>();
}
}
return list;
}