Create two_sum_in_bst_iv.cpp

Author: avidLearnerInProgress

two_sum_in_bst_iv.cpp

@@ -0,0 +1,58 @@
+/**
+ * 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 BSTIterator {
+        
+    stack<TreeNode*> st;
+    bool reverse;
+    
+    void partialInorder(TreeNode *root) {
+        while(root) {
+            st.push(root);
+            root = !reverse ? root->left : root->right;
+        }
+    }
+    public:    
+        BSTIterator(TreeNode *root, bool reverse = false) : reverse(reverse) {
+            partialInorder(root);
+        }
+        
+        int next() {
+            auto next = st.top();
+            st.pop();
+            
+            auto itr = !reverse ? next->right : next->left; //for leftItr(check right subtree) and inverse for other 
+            if(itr) {
+                partialInorder(itr);
+            }
+            return next->val;
+        }
+};
+
+/*based on approach mentioned here - https://leetcode.com/problems/two-sum-iv-input-is-a-bst/discuss/1420711*/
+/*TC: O(n) | SC: O(h)*/
+class Solution {
+public:
+    bool findTarget(TreeNode* root, int k) {
+        BSTIterator leftItr(root), rightItr(root, true);
+        int left = leftItr.next(), right = rightItr.next();
+        
+        while(left < right) {
+            if(left + right == k) return true;
+            if(left + right < k)
+                left = leftItr.next();
+            else
+                right = rightItr.next();
+        }
+        return false;
+    }
+};