107. 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,#,#,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.
链接: http://leetcode.com/problems/binary-tree-level-order-traversal-ii/
一刷,比102题只多了最后一步,还是用了2个list,第二遍可以试一下DFS
1 class Solution(object): 2 def levelOrderBottom(self, root): 3 if not root: 4 return [] 5 result = [] 6 prev = [root] 7 cur = [] 8 9 while prev: 10 for elem in prev: 11 if elem.left: 12 cur.append(elem.left) 13 if elem.right: 14 cur.append(elem.right) 15 result.append([e.val for e in prev]) 16 prev, cur = cur, [] 17 result.reverse() 18 return result
2/16/2017, Java
错误长的都不知道该怎么写了,而且这段代码效率低得可以不看了
1. List, ArrayList, LinkedList, Queue, Stack的各种区别
2. Interface, Implementing Class的区别,关系
3. List<List<Integer>> ret = new ArrayList<>(); 语法上这是怎么一回事
4. stack(level); level.clear();之后stack里面都是空List,这让用Python的人怎么活。所以Java是怎么存储数据的!
5. 效率提高
1 public class Solution { 2 public List<List<Integer>> levelOrderBottom(TreeNode root) { 3 4 Queue<TreeNode> current = new LinkedList<TreeNode>(); 5 TreeNode p = new TreeNode(-1); 6 List<List<Integer>> ret = new ArrayList<>(); 7 List<Integer> level = new ArrayList<Integer>(); 8 Stack<List<Integer>> stack = new Stack<>(); 9 int arraySize = 0; 10 11 if (root == null) return ret; 12 13 current.add(root); 14 while(current != null && !current.isEmpty()) { 15 arraySize = current.size(); 16 for(int i = 0; i < arraySize; i++) { 17 p = current.poll(); 18 if (p.left != null) { 19 current.add(p.left); 20 } 21 if (p.right != null) { 22 current.add(p.right); 23 } 24 level.add(p.val); 25 } 26 stack.push(level); 27 level = new ArrayList<Integer>(); 28 } 29 while(!stack.empty()) { 30 ret.add(stack.pop()); 31 } 32 return ret; 33 } 34 }