LeetCode 103. Binary tree zigzag level order transverse - Binary Tree Series Question 5

Given the root of a binary tree, return the zigzag level order traversal of its nodes' values. (i.e., from left to right, then right to left for the next level and alternate between).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[20,9],[15,7]]

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -100 <= Node.val <= 100

The title requires level Z Traverse the whole binary tree in font order , That is, in behavioral units , A line from left to right , The next line is from right to left , In this way, the binary tree is traversed alternately .

If you are familiar with the horizontal traversal binary tree algorithm , This question is relatively simple . I think about it , utilize BFS The algorithm traverses in behavioral units . Define a queue queue, Put the whole row of nodes into the queue each time , In each cycle, all nodes in the current queue （ That is, the nodes of the whole line ） Take out one by one , Put the values into a temporary array , Then take out the left and right child nodes of the node （ If it exists ） Put it in the queue . As required Z Font type , Every other row needs to reverse the temporary array . Finally, add all the numbers of the temporary array to the answer array .

class Solution:
    def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        
        res = []
        l2r = True
        q = deque([root])
        
        while q:
            n = len(q)
            level = []
            for i in range(n):
                node = q.popleft()
                level.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            
            if l2r:
                l2r = False
            else:
                l2r = True
                level.reverse()
            res.append(level)
            
        return res