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。

 1 class Solution {
 2 public:
 3     bool isValidSerialization(string preorder) {
 4         istringstream iss(preorder);
 5         string node;
 6         stack<bool> leftPathTrack;
 7         bool tracked = false;
 8         while (getline(iss, node, ‘,‘)) {
 9             if (tracked && leftPathTrack.empty()) return false;
10             tracked = true;
11             if (node == "#") {
12                 //for the case the root node is ‘#‘
13                 if (leftPathTrack.empty()) continue;
14                 /*
15                 If the current node is left child, pop it and mark the next node as right.
16                 Else, pop all the nodes in the stack until the top node is a left child,
17                 pop it and mark the next node as right.
18                 */
19                 while (!leftPathTrack.empty() && !leftPathTrack.top())
20                     leftPathTrack.pop();
21                 if (!leftPathTrack.empty()) {
22                     leftPathTrack.pop();
23                     leftPathTrack.push(false);
24                 }
25             } else {
26                 leftPathTrack.push(true);
27             }
28         }
29         return leftPathTrack.empty();
30     }
31 };