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;
	}
}