原题链接在这里:https://leetcode.com/problems/binary-tree-level-order-traversal-ii/
题目:
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] ]
题解:
与Binary Tree Level Order Traversal相似,只是返过来加链表。每次把item从头加进res中.
Time Complexity: O(n). Space O(n).
AC Java:
1 /** 2 * Definition for a binary tree node. 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 17 LinkedList<TreeNode> que = new LinkedList<TreeNode>(); 18 que.add(root); 19 int curCount = 1; 20 int nextCount = 0; 21 List<Integer> item = new ArrayList<Integer>(); 22 while(!que.isEmpty()){ 23 TreeNode cur = que.poll(); 24 curCount--; 25 item.add(cur.val); 26 27 if(cur.left != null){ 28 que.add(cur.left); 29 nextCount++; 30 } 31 if(cur.right != null){ 32 que.add(cur.right); 33 nextCount++; 34 } 35 if(curCount == 0){ 36 res.add(0, new ArrayList<Integer>(item)); 37 item = new ArrayList<Integer>(); 38 curCount = nextCount; 39 nextCount = 0; 40 } 41 } 42 return res; 43 } 44 }