Java [Leetcode 107]Binary Tree Level Order Traversal II

题目描述：

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]
]

解题思路：

运用广度优先搜索方法，运用队列的方法，每次遍历一层的叶节点，并把下一层的叶节点加入到队列中。每次遍历一层节点结束时候，将该层节点组成的list放到整体的list之前，实现由底到高的排列。

代码如下：

/**
* 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>> list = new LinkedList<List<Integer>>();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
if(root == null)
return list;
queue.offer(root);
while(!queue.isEmpty()){
int num = queue.size();
List<Integer> levelList = new LinkedList<Integer>();
for(int i = 0; i < num; i++){
if(queue.peek().left != null)
queue.offer(queue.peek().left);
if(queue.peek().right != null)
queue.offer(queue.peek().right);
levelList.add(queue.poll().val);
}
list.add(0, levelList);
}
return list;
}
}