Python BFS solution with clear comments

  • 0
    class Solution(object):
        def zigzagLevelOrder(self, root):
            :type root: TreeNode
            :rtype: List[List[int]]
            res = [] # result list
            cnt = 0 # use to toggle printed order
            if not root:
                return res
            now = [root]
            while now:
                cnt += 1
                tmp = [] # next level elements
                vals = [] # current level's node value
                for x in now:
                    if x.left:
                    if x.right:
                # zigzag
                if cnt % 2 == 1:
                    res.append([v for v in vals])
                    res.append([v for v in reversed(vals)])
                # progress level
                now = tmp
            return res

    The idea is simple. I use a counter to determine whether to print the current level in normal order or reversed order. And be sure to put node.val instead of node to the final result.

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