|
| 1 | +packageeasy; |
| 2 | + |
| 3 | +importclasses.TreeNode; |
| 4 | + |
| 5 | +importjava.util.ArrayList; |
| 6 | +importjava.util.Collections; |
| 7 | +importjava.util.LinkedList; |
| 8 | +importjava.util.List; |
| 9 | +importjava.util.Queue; |
| 10 | + |
| 11 | +/**107. Binary Tree Level Order Traversal II |
| 12 | +
|
| 13 | + Total Accepted: 91577 |
| 14 | + Total Submissions: 258036 |
| 15 | + Difficulty: Easy |
| 16 | +
|
| 17 | +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). |
| 18 | +
|
| 19 | +For example: |
| 20 | +Given binary tree [3,9,20,null,null,15,7], |
| 21 | +
|
| 22 | + 3 |
| 23 | + / \ |
| 24 | + 9 20 |
| 25 | + / \ |
| 26 | + 15 7 |
| 27 | +
|
| 28 | +return its bottom-up level order traversal as: |
| 29 | +
|
| 30 | +[ |
| 31 | + [15,7], |
| 32 | + [9,20], |
| 33 | + [3] |
| 34 | +] |
| 35 | +
|
| 36 | +*/ |
| 37 | +publicclassBinaryTreeLevelOrderTraversalII { |
| 38 | +publicList<List<Integer>>levelOrder(TreeNoderoot) { |
| 39 | +List<List<Integer>>result =newArrayList<List<Integer>>(); |
| 40 | +if(root ==null)returnresult; |
| 41 | + |
| 42 | +Queue<TreeNode>q =newLinkedList<TreeNode>(); |
| 43 | +q.offer(root); |
| 44 | +while(!q.isEmpty()){ |
| 45 | +List<Integer>thisLevel =newArrayList<Integer>(); |
| 46 | +intqSize =q.size(); |
| 47 | +for(inti =0;i <qSize;i++){ |
| 48 | +TreeNodecurr =q.poll(); |
| 49 | +thisLevel.add(curr.val); |
| 50 | +if(curr.left !=null)q.offer(curr.left); |
| 51 | +if(curr.right !=null)q.offer(curr.right); |
| 52 | +} |
| 53 | +result.add(thisLevel); |
| 54 | +} |
| 55 | +Collections.reverse(result);//this is the only line that gets added/changed to the previous solution. |
| 56 | +returnresult; |
| 57 | +} |
| 58 | +} |