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:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        
        LinkedList<TreeNode> que = new LinkedList<>();
        que.add(root);
        boolean reverse = false;
        
        while(!que.isEmpty()){
            int size = que.size();
            List<Integer> item = new ArrayList<>();
            while(size-- > 0){
                TreeNode cur = que.poll();
                item.add(cur.val);
                if(cur.left != null){
                    que.add(cur.left);
                }
                
                if(cur.right != null){
                    que.add(cur.right);
                }
            }
            
            if(reverse){
               Collections.reverse(item); 
            }
            
            res.add(item);
            reverse = !reverse;
        }
        
        return res;
    }
}

AC C++:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        if(!root){
            return {};
        }

        vector<vector<int>> res;
        deque<TreeNode*> que;
        que.push_back(root);
        bool shouldReverse = false;
        while(!que.empty()){
            vector<int> item;
            int size = que.size();
            while(size-- > 0){
                TreeNode* cur = que.front();
                que.pop_front();
                item.push_back(cur->val);

                if(cur->left){
                    que.push_back(cur->left);
                }

                if(cur->right){
                    que.push_back(cur->right);
                }
            }

            if(shouldReverse){
                reverse(item.begin(), item.end());
            }

            shouldReverse = !shouldReverse;
            res.push_back(item);
        }

        return res;
    }
};

AC Python:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        
        res = []
        que = deque([root])
        should_reverse = False

        while que:
            item = []
            size = len(que)

            for _ in range(size):
                cur = que.popleft()
                item.append(cur.val)

                if cur.left:
                    que.append(cur.left)
                if cur.right:
                    que.append(cur.right)

            if should_reverse:
                item.reverse()

            should_reverse = not should_reverse
            res.append(item)

        return res