[103]二叉树的锯齿形层次遍历.py

# 给定一个二叉树，返回其节点值的锯齿形层次遍历。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。 
# 
#  例如： 
# 给定二叉树 [3,9,20,null,null,15,7], 
# 
#      3
#    / \
#   9  20
#     /  \
#    15   7
#  
# 
#  返回锯齿形层次遍历如下： 
# 
#  [
#   [3],
#   [20,9],
#   [15,7]
# ]
#  
#  Related Topics 栈 树 广度优先搜索 
#  👍 270 👎 0


# leetcode submit region begin(Prohibit modification and deletion)
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]:
        # 方式1：和层次遍历一样，遇到偶数行反转
        if not root: return []
        res, cur_level = [], [root]
        depth = 0

        while cur_level:
            tmp, next_level = [], []

            for node in cur_level:
                tmp.append(node.val)
                if node.left:
                    next_level.append(node.left)
                if node.right:
                    next_level.append(node.right)
            if depth % 2 == 1:
                res.append(tmp[::-1])
            else:
                res.append(tmp)
            depth += 1
            cur_level = next_level
        return res
    def zigzagLevelOrder1(self, root: TreeNode) -> List[List[int]]:
        # 方式2：dfs
        res = []
        def helper(root, depth):
            if not root: return
            if len(res) == depth:
                res.append([])
            if depth % 2 == 0:
                res[depth].append(root.val)
            else:
                res[depth].insert(0, root.val)
            helper(root.left, depth + 1)
            helper(root.right, depth + 1)

        helper(root, 0)
        return res