LeetCode@107_Binary_Tree_Level_Order_Traversal_II

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]
]

java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {

Queue<TreeNode> queue = new LinkedList<>();
List<Integer> treeVal = new LinkedList<>();
List<List<Integer>> res = new LinkedList<>();

public List<List<Integer>> levelOrderBottom(TreeNode root)
{
if (root == null)
return res;
queue.add(root);
treeVal.add(root.val);
res.add(treeVal);
while (!queue.isEmpty())
{
List<Integer> currVal = new LinkedList<>();

//需要判断有多少个subtreeval e.t level2 有两个数值【2，3】
//#############
int len = queue.size();
for(int i =0;i<len;i++)
{
TreeNode currNode = queue.poll();
if (currNode.left != null)
{
currVal.add(currNode.left.val);
queue.add(currNode.left);
}
if (currNode.right != null)
{
currVal.add(currNode.right.val);
queue.add(currNode.right);
}
}
if (!currVal.isEmpty())
{
res.add(currVal);
}

}
Collections.reverse(res);
return res;
}
}