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;}}