Create Day 24 Construct Binary Search Tree from Preorder Traversal.cpp

Author: tusharjaiswal123

Day 24 Construct Binary Search Tree from Preorder Traversal.cpp

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+PROBLEM:
+
+
+
+Return the root node of a binary search tree that matches the given preorder traversal.
+
+(Recall that a binary search tree is a binary tree where for every node, any descendant of node.left has a value < node.val,
+and any descendant of node.right has a value > node.val.  Also recall that a preorder traversal displays the value of the
+node first, then traverses node.left, then traverses node.right.)
+
+It's guaranteed that for the given test cases there is always possible to find a binary search tree with the given requirements.
+
+Example 1:
+
+Input: [8,5,1,7,10,12]
+Output: [8,5,10,1,7,null,12]
+
+
+
+
+SOLUTION:
+
+
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    TreeNode* bstFromPreorder(vector<int>& preorder) {
+        int i=0;
+        return bst(preorder,INT_MAX,i);
+        
+    }
+    
+    TreeNode* bst(vector<int>& preorder,int upper,int &i)
+    {
+        if(i==preorder.size() || preorder[i] > upper)
+            return NULL;
+        
+        TreeNode* newnode = new TreeNode(preorder[i]);
+        i++;
+        newnode->left = bst(preorder,newnode->val,i);
+        newnode->right = bst(preorder,upper,i);
+        
+        return newnode;
+    }
+};