[LeetCode] 103. Binary Tree Zigzag Level Order Traversal _ Medium tag: BFS

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]
]

这个题目的思路也是典型的BFS, 只是每一层分别正序和倒序, 我还是利用102. Binary Tree Level Order Traversal_Medium tag: BFS的思路,只是多加了一个len(ans)%2== 0 去判断是应该正序还是倒序, 如果是倒序, 就把temp reverse然后append进入ans里面即可.

1. Constraints

1) edge case, root is None

2. Ideas

BFS:     T: O(n)   S: O(n)

3.code:

 class Solution:
     def zigzagLevel(self, root):
         ans = []
         if not root: return ans # edge case
         queue = collections.deque([root])
         while queue:
             size, level = len(queue), []
             for _ in range(size):
                 node = queue.popleft()
                 level.append(node.val)
                 if node.left:
                     queue.append(node.left)
                 if node.right:
                     queue.append(node.right)
             if len(ans)%2 ==0:
                 ans.append(level)
             else:
                 ans.append(level[::-1])
         return ans