LeetCode 103. Binary Tree Zigzag Level Order Traversal
原题链接在这里：https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/
题目：
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7] ]
题解：
这道题是BFS的变形，与Binary Tree Level Order Traversal相似。但是要求偶数行从左到右，奇数行从右到左。
其实还是BFS, 只不过需要添加一个flag来表明是否需要reverse, 这里用boolean reverse 表示.
Time Complexity: O(n).
Space: O(n). que最多有n/2个节点。
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 class Solution { 11 public List<List<Integer>> zigzagLevelOrder(TreeNode root) { 12 List<List<Integer>> res = new ArrayList<>(); 13 if(root == null){ 14 return res; 15 } 16 17 LinkedList<TreeNode> que = new LinkedList<>(); 18 que.add(root); 19 boolean reverse = false; 20 21 while(!que.isEmpty()){ 22 int size = que.size(); 23 List<Integer> item = new ArrayList<>(); 24 while(size-- > 0){ 25 TreeNode cur = que.poll(); 26 item.add(cur.val); 27 if(cur.left != null){ 28 que.add(cur.left); 29 } 30 31 if(cur.right != null){ 32 que.add(cur.right); 33 } 34 } 35 36 if(reverse){ 37 Collections.reverse(item); 38 } 39 40 res.add(item); 41 reverse = !reverse; 42 } 43 44 return res; 45 } 46 }
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>> zigzagLevelOrder(TreeNode* root) { 15 if(!root){ 16 return {}; 17 } 18 19 vector<vector<int>> res; 20 deque<TreeNode*> que; 21 que.push_back(root); 22 bool shouldReverse = false; 23 while(!que.empty()){ 24 vector<int> item; 25 int size = que.size(); 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 if(shouldReverse){ 41 reverse(item.begin(), item.end()); 42 } 43 44 shouldReverse = !shouldReverse; 45 res.push_back(item); 46 } 47 48 return res; 49 } 50 };
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 zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: 9 if not root: 10 return [] 11 12 res = [] 13 que = deque([root]) 14 should_reverse = False 15 16 while que: 17 item = [] 18 size = len(que) 19 20 for _ in range(size): 21 cur = que.popleft() 22 item.append(cur.val) 23 24 if cur.left: 25 que.append(cur.left) 26 if cur.right: 27 que.append(cur.right) 28 29 if should_reverse: 30 item.reverse() 31 32 should_reverse = not should_reverse 33 res.append(item) 34 35 return res 36 37
