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] ]
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
Solution:
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 List<List<Integer>> levelOrderBottom(TreeNode root) { 12 List<List<Integer>> res = new ArrayList<List<Integer>>(); 13 if (root==null) 14 return res; 15 16 //Visit all node by using BFS. 17 List<TreeNode> queue = new ArrayList<TreeNode>(); 18 List<Integer> depth = new ArrayList<Integer>(); 19 queue.add(root); 20 depth.add(1); 21 int index = 0; 22 int curDepth = -1; 23 TreeNode curNode = null; 24 while (index<queue.size()){ 25 curNode = queue.get(index); 26 curDepth = depth.get(index); 27 if (curNode.left!=null){ 28 queue.add(curNode.left); 29 depth.add(curDepth+1); 30 } 31 32 if (curNode.right!=null){ 33 queue.add(curNode.right); 34 depth.add(curDepth+1); 35 } 36 37 index++; 38 } 39 40 //Get the max depth, which is the number of lists in the result. 41 int maxDepth = depth.get(depth.size()-1); 42 for (int i=0;i<maxDepth;i++) 43 res.add(new ArrayList<Integer>()); 44 45 //Put the value of each node into the corresponding list in the result. 46 for (int i=0;i<queue.size();i++){ 47 curNode = queue.get(i); 48 curDepth = depth.get(i); 49 index = maxDepth-curDepth; 50 List<Integer> tempList = res.get(index); 51 tempList.add(curNode.val); 52 } 53 54 return res; 55 } 56 }
Using BFS to visit and record every tree node and its depth. Then put the value in each tree node into corresponding level list.