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_
    /   \

  / \   / \
      #
/ \ / \   / \
# # # #   # #

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();
     }
 };