feat: solve construct-binary-tree-from-preorder-and-inorder-traversal

Author: lylaminju

construct-binary-tree-from-preorder-and-inorder-traversal/pmjuu.py

@@ -0,0 +1,36 @@
+from typing import List, Optional
+
+# Definition for a binary tree node.
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+class Solution:
+    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
+        inorder_map = {value: idx for idx, value in enumerate(inorder)}
+        self.preorder_idx = 0
+
+        def helper(left: int, right: int) -> Optional[TreeNode]:
+            if left > right:
+                return None
+            
+            root_val = preorder[self.preorder_idx]
+            self.preorder_idx += 1
+
+            root = TreeNode(root_val)
+            root.left = helper(left, inorder_map[root_val] - 1)
+            root.right = helper(inorder_map[root_val] + 1, right)
+
+            return root
+        
+        return helper(0, len(inorder) - 1)
+
+
+# Time Complexity: O(n)
+# - Each node is visited exactly once in preorder, and the dictionary lookup in inorder_map takes O(1) per node.
+
+# Space Complexity: O(n)
+# - The hash map (inorder_map) uses O(n) space.
+# - The recursion stack uses up to O(h) space, where h is the height of the tree (O(n) in the worst case, O(log n) for a balanced tree).