【leetcode】331. Verify Preorder Serialization of a Binary Tree

题目如下：

One way to serialize a binary tree is to use pre-order traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as #.
     _9_

    /   \

   3     2

  / \   / \

 4   1  #  6

/ \ / \   / \

# # # #   # #
For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where # represents a null node.

Given a string of comma separated values, verify whether it is a correct preorder traversal serialization of a binary tree. Find an algorithm without reconstructing the tree.

Each comma separated value in the string must be either an integer or a character '#' representing null pointer.

You may assume that the input format is always valid, for example it could never contain two consecutive commas such as "1,,3".

Example 1:
Input: "9,3,4,#,#,1,#,#,2,#,6,#,#"

Output: true
Example 2:
Input: "1,#"

Output: false
Example 3:
Input: "9,#,#,1"

Output: false

解题思路：我的方法是模拟树的遍历，依次pop出preorder中的节点存入queue中，用direction记录遍历的方向。direction=1表示往左子树方向的，等于2表示往右子树方向，遇到#号则变换一次方向。如果preorder的顺序是合法的，那么最后的结果一定是preorder和queue里面都是空。

代码如下：

class Solution(object):

    def isValidSerialization(self, preorder):

        """

        :type preorder: str

        :rtype: bool

        """

        if preorder == '#':

            return True

        elif preorder[0] == '#':

            return False

        preorder = preorder.split(',')

        direction = 1

        queue = [int(preorder.pop(0))]

        while len(queue) > 0 and len(preorder) > 0:

            item = preorder.pop(0)

            if item == '#':

                direction += 1

                if direction % 3 == 0:

                    direction -= 1

                    queue.pop(-1)

                continue

            queue.append(item)

        return len(preorder) == 0 and len(queue) == 0

巴特西

【leetcode】331. Verify Preorder Serialization of a Binary Tree

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