New Problem Solution "Verify Preorder Serialization of a Binary Tree"

README.md

@@ -54,6 +54,7 @@ LeetCode
 |338|[Counting Bits](https://leetcode.com/problems/counting-bits/) | [C++](./algorithms/cpp/countingBits/CountingBits.cpp)|Medium|
 |337|[House Robber III](https://leetcode.com/problems/house-robber-iii/) | [C++](./algorithms/cpp/houseRobber/houseRobberIII.cpp)|Medium|
 |334|[Increasing Triplet Subsequence](https://leetcode.com/problems/increasing-triplet-subsequence/) | [C++](./algorithms/cpp/increasingTripletSubsequence/increasingTripletSubsequence.cpp)|Medium|
+|331|[Verify Preorder Serialization of a Binary Tree](https://leetcode.com/problems/verify-preorder-serialization-of-a-binary-tree/) | [C++](./algorithms/cpp/verifyPreorderSerializationOfABinaryTree/VerifyPreorderSerializationOfABinaryTree.cpp)|Medium|
 |330|[Patching Array](https://leetcode.com/problems/patching-array/) | [C++](./algorithms/cpp/patchingArray/PatchingArray.cpp)|Medium|
 |329|[Longest Increasing Path in a Matrix](https://leetcode.com/problems/longest-increasing-path-in-a-matrix/) | [C++](./algorithms/cpp/longestIncreasingPathInAMatrix/LongestIncreasingPathInAMatrix.cpp)|Medium|
 |328|[Odd Even Linked List](https://leetcode.com/problems/odd-even-linked-list/) | [C++](./algorithms/cpp/oddEvenLinkedList/OddEvenLinkedList.cpp)|Easy|

algorithms/cpp/verifyPreorderSerializationOfABinaryTree/VerifyPreorderSerializationOfABinaryTree.cpp

@@ -0,0 +1,85 @@
+// Source : https://leetcode.com/problems/verify-preorder-serialization-of-a-binary-tree/
+// Author : Hao Chen
+// Date   : 2017-01-06
+
+/*************************************************************************************** 
+ *
+ * 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.
+ ***************************************************************************************/
+
+class Solution {
+public:
+
+    // we know the following facts:
+    //   1) if we met a non-null node, then this node will generate two child node.
+    //   2) if we met a null node, then this node will generate zero child node.
+    //
+    // so the idea is, 
+    //   1) we can have a counter to calculate how many node we are going to expect 
+    //   2) once we have the expected node, then we can decrease the counter.
+    //   3) finally, we will check the counter is zero or not.
+    //
+    // the algorithm as below:
+    //   1) when we meet a non-null node, just simply do `counter++`. because:
+    //      1.1) we will expect 2 more node after, then we do `counter += 2`. 
+    //      1.2) but the current node also meet the expection of parent node , so, it need remove 1 in counter.
+    //           finally, the `counter = counbter + 2 -1`
+    //   2) when we meet a null node, just simply do `counter--`.
+    
+    bool isValidSerialization(string preorder) {
+        vector<string> list;
+        split(preorder, ',', list);
+        //we initailize the counter as 1, 
+        //because we expect at lease 1 node in the tree.
+        int node_expected = 1;
+        for (auto node : list) {
+            if (node_expected == 0) return false;
+            node == "#" ? node_expected-- : node_expected++;
+        }
+        return node_expected == 0;
+    }
+    
+    void split(const string &s, char delim, vector<string> &elems) {
+        stringstream ss(s);
+        string item;
+        while (getline(ss, item, delim)) {
+            elems.push_back(item);
+        }
+    }
+};