Create 889. Construct Binary Tree from Preorder and Postorder Traversal (#724)

Author: Chayandas07

889. Construct Binary Tree from Preorder and Postorder Traversal

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+class Solution {
+    int preIndex = 0;
+    unordered_map<int, int> postMap;
+    
+    TreeNode* construct(vector<int>& preorder, vector<int>& postorder, int postStart, int postEnd) {
+        // Base case: if no more nodes to process
+        if (postStart > postEnd || preIndex >= preorder.size()) {
+            return nullptr;
+        }
+        
+        // Create root node from current preorder value
+        TreeNode* root = new TreeNode(preorder[preIndex++]);
+        
+        // If we've processed all nodes or this is a leaf node
+        if (postStart == postEnd) {
+            return root;
+        }
+        
+        // Find the index of next preorder value in postorder
+        // This will be the end of left subtree in postorder
+        int postIndex = postMap[preorder[preIndex]];
+        
+        // Only proceed if we found a valid index within our current range
+        if (postIndex >= postStart && postIndex <= postEnd) {
+            // Recursively construct left subtree
+            // postStart to postIndex is the left subtree in postorder
+            root->left = construct(preorder, postorder, postStart, postIndex);
+            
+            // Recursively construct right subtree
+            // postIndex + 1 to postEnd - 1 is the right subtree in postorder
+            root->right = construct(preorder, postorder, postIndex + 1, postEnd - 1);
+        }
+        
+        return root;
+    }
+    
+public:
+    TreeNode* constructFromPrePost(vector<int>& preorder, vector<int>& postorder) {
+        int n = postorder.size();
+        
+        // Create map of value to index for postorder array
+        for (int i = 0; i < n; i++) {
+            postMap[postorder[i]] = i;
+        }
+        
+        return construct(preorder, postorder, 0, n - 1);
+    }
+};