331. Verify Preorder Serialization of a Binary Tree-优快云博客

本文介绍了一种不重建树即可验证字符串是否为正确二叉树前序遍历序列的方法。通过栈操作来检查字符串中节点与空指针的合法性。

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Problem：

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

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

Solution：

class Solution {
public:
bool isValidSerialization(string preorder) {
stack<char> a;
for (int i = 0; i < preorder.size(); i++)
{
if (preorder[i] == ',')
{
continue;
}
if (preorder[i] != '#')
{
a.push(preorder[i]);
}
else
{
while (!a.empty())
{
if (a.top() != '#')
{
break;
}
a.pop();
if (a.top() == '#' || a.empty())
{
return false;
}
else
{
a.pop();
}

}
a.push('#');
}
}
if (a.top() == '#' && a.size() == 1)
{
return true;
}
else
{
return false;
}
}
};