LeetCode Verify Preorder Serialization of a Binary Tree

Description

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

The original problem is here.

My Solution

I solve this problem in C++, as below:

class Solution {
public:
    bool isValidSerialization(string preorder) {
        stack<char> nodes;
        for(int i = 0, dotCount = 0; i < preorder.length(); i++) {
            if(dotCount % 2 == 1) {
                if (preorder[i] != ',') {
                    continue;
                } else {
                    dotCount ++;
                }
                continue;
            } else {
                dotCount ++;
            }
            nodes.push(preorder[i]);
        }
        int symbolCount = 0;
        while(!nodes.empty() && (nodes.size() + symbolCount) >= 3) {
            if (nodes.top() == '#') {
                symbolCount ++;
                nodes.pop();
            } else {
                if (symbolCount >= 2) {
                    nodes.pop();
                    nodes.push('#');
                    symbolCount -= 2;
                } else {
                    return false;
                }
            }
        }
        if ( nodes.size() == 1 && nodes.top() == '#' && symbolCount == 0) {
            return true;
        }
        return false;
    }
};

Note

To solve the problem, use a stack to save the node and verify every child.