Verify Preorder Serialization of a Binary Tree -- LeetCode

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：将字符串按照逗号分隔，然后依次读入节点。用一个栈来记录当前的节点是左孩子还是右孩子。

class Solution {
public:
    bool isValidSerialization(string preorder) {
        istringstream iss(preorder);
        string node;
        stack<bool> leftPathTrack;
        bool tracked = false;
        while (getline(iss, node, ',')) {
            if (tracked && leftPathTrack.empty()) return false;
            tracked = true;
            if (node == "#") {
                //for the case the root node is '#'
                if (leftPathTrack.empty()) continue;
                /*
                If the current node is left child, pop it and mark the next node as right.
                Else, pop all the nodes in the stack until the top node is a left child,
                pop it and mark the next node as right.
                */
                while (!leftPathTrack.empty() && !leftPathTrack.top()) 
                    leftPathTrack.pop();
                if (!leftPathTrack.empty()) {
                    leftPathTrack.pop();
                    leftPathTrack.push(false);
                }
            } else {
                leftPathTrack.push(true);
            }
        }
        return leftPathTrack.empty();
    }
};