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Author: coderpheonix

1028. Recover a Tree From Preorder Traversal.cpp

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+/**
+ * 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* recoverFromPreorder(string traversal) {
+        stack<TreeNode*> st;
+        int i = 0, n = traversal.size();
+
+        while (i < n) {
+            int depth = 0;
+            // Count the number of dashes to determine depth
+            while (i < n && traversal[i] == '-') {
+                depth++;
+                i++;
+            }
+
+            // Extract the node value
+            int value = 0;
+            while (i < n && isdigit(traversal[i])) {
+                value = value * 10 + (traversal[i] - '0');
+                i++;
+            }
+
+            // Create a new node
+            TreeNode* node = new TreeNode(value);
+
+            // If the stack size is greater than depth, pop elements to find the correct parent
+            while (st.size() > depth) {
+                st.pop();
+            }
+
+            // Attach the node to its parent
+            if (!st.empty()) {
+                if (!st.top()->left) {
+                    st.top()->left = node;
+                } else {
+                    st.top()->right = node;
+                }
+            }
+
+            // Push the node to the stack
+            st.push(node);
+        }
+
+        // The root node is the bottom-most element in the stack
+        while (st.size() > 1) {
+            st.pop();
+        }
+
+        return st.top();
+    }
+};