Leetcode: Binary Tree level order transversal 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,#,#,15,7},    3   /   9  20    /     15   7return its bottom-up level order traversal as:[  [15,7],  [9,20],  [3]]

Difficulty based on binary tree level order transversal: 20. You only need to make an inverted order for the final result. The format is collections. Reverse (list <?> List)

 1 /** 2  * Definition for binary tree 3  * public class TreeNode { 4  *     int val; 5  *     TreeNode left; 6  *     TreeNode right; 7  *     TreeNode(int x) { val = x; } 8  * } 9  */10 public class Solution {11     public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {12         ArrayList<ArrayList<Integer>> lists = new ArrayList<ArrayList<Integer>> ();13         if (root == null) return lists;14         LinkedList<TreeNode> queue = new LinkedList<TreeNode>();15         queue.add(root);16         int ParentNumInQ = 1;17         int ChildNumInQ = 0;18         ArrayList<Integer> list = new ArrayList<Integer>();19         while (!queue.isEmpty()) {20             TreeNode cur = queue.poll();21             list.add(cur.val);22             ParentNumInQ--;23             if (cur.left != null) {24                 queue.add(cur.left);25                 ChildNumInQ++;26             }27             if (cur.right != null) {28                 queue.add(cur.right);29                 ChildNumInQ++;30             }31             if (ParentNumInQ == 0) {32                 ParentNumInQ = ChildNumInQ;33                 ChildNumInQ = 0;34                 lists.add(list);35                 list = new ArrayList<Integer>();36             }37         }38         Collections.reverse(lists);39         return lists;40     }41 }

 

Leetcode: Binary Tree level order transversal II