leetcode Verify Preorder Serialization of a Binary Tree

最新推荐文章于 2024-08-26 07:28:39 发布

hr_whisper

最新推荐文章于 2024-08-26 07:28:39 发布

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本文提供了一种不重建树即可验证二叉树序列化的算法。通过分析预序遍历字符串，利用栈来检查节点的有效性，确保序列化正确无误。

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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:
"9,3,4,#,#,1,#,#,2,#,6,#,#"
Return true

Example 2:
"1,#"
Return false

Example 3:
"9,#,#,1"

Return false

思路：

栈
入度和出度的差

原文+解释： https://www.hrwhisper.me/leetcode-verify-preorder-serialization-of-a-binary-tree/

栈

class Solution(object):
    def isValidSerialization(self, preorder):
        """
        :type preorder: str
        :rtype: bool
        """
        stack = []
        for x in preorder.split(','):
            stack += [x]
            while len(stack) >= 3 and stack[-2:] == ['#','#'] and stack[-3]!='#':
                stack = stack[:-3] +['#'] # pop *3 and append #
        return len(stack) == 1 and stack[0] == '#'

public boolean isValidSerialization(String preorder) {
    String[] nodes = preorder.split(",");
    int diff = 1;
    for (String node: nodes) {
        if (--diff < 0) return false;
        if (!node.equals("#")) diff += 2;
    }
    return diff == 0;
}

本文是leetcode 331 Verify Preorder Serialization of a Binary Tree 的题解，

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