LeetCode 107. Binary Tree Level Order Traversal II
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() {} 8 * TreeNode(int val) { this.val = val; } 9 * TreeNode(int val, TreeNode left, TreeNode right) { 10 * this.val = val; 11 * this.left = left; 12 * this.right = right; 13 * } 14 * } 15 */ 16 class Solution { 17 public List<List<Integer>> levelOrderBottom(TreeNode root) { 18 LinkedList<List<Integer>> res = new LinkedList<>(); 19 if(root == null){ 20 return res; 21 } 22 23 LinkedList<TreeNode> que = new LinkedList<>(); 24 que.add(root); 25 while(!que.isEmpty()){ 26 int size = que.size(); 27 List<Integer> item = new ArrayList<>(); 28 while(size-- > 0){ 29 TreeNode cur = que.poll(); 30 item.add(cur.val); 31 if(cur.left != null){ 32 que.add(cur.left); 33 } 34 35 if(cur.right != null){ 36 que.add(cur.right); 37 } 38 } 39 40 res.addFirst(item); 41 } 42 43 return res; 44 } 45 }
AC C++:
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode() : val(0), left(nullptr), right(nullptr) {} 8 * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} 9 * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} 10 * }; 11 */ 12 class Solution { 13 public: 14 vector<vector<int>> levelOrderBottom(TreeNode* root) { 15 if(!root){ 16 return {}; 17 } 18 19 vector<vector<int>> res; 20 deque<TreeNode*> que; 21 que.push_back(root); 22 23 while(!que.empty()){ 24 int size = que.size(); 25 vector<int> item; 26 while(size-- > 0){ 27 TreeNode* cur = que.front(); 28 que.pop_front(); 29 item.push_back(cur->val); 30 31 if(cur->left){ 32 que.push_back(cur->left); 33 } 34 35 if(cur->right){ 36 que.push_back(cur->right); 37 } 38 } 39 40 res.push_back(item); 41 } 42 43 reverse(res.begin(), res.end()); 44 return res; 45 } 46 };
AC Python:
1 # Definition for a binary tree node. 2 # class TreeNode: 3 # def __init__(self, val=0, left=None, right=None): 4 # self.val = val 5 # self.left = left 6 # self.right = right 7 class Solution: 8 def levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]: 9 if not root: 10 return [] 11 12 res = deque() 13 que = deque([root]) 14 15 while que: 16 size = len(que) 17 item = [] 18 19 for _ in range(size): 20 cur = que.popleft() 21 item.append(cur.val) 22 23 if cur.left: 24 que.append(cur.left) 25 26 if cur.right: 27 que.append(cur.right) 28 29 res.appendleft(item) 30 31 return list(res) 32
