solve: construct binary tree from preorder and inorder traversal

Author: sounmind

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

@@ -0,0 +1,37 @@
+from typing import List, Optional
+
+
+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]:
+        if not preorder or not inorder:
+            return None
+
+        root_val = preorder.pop(0)
+
+        root_inorder_index = inorder.index(root_val)
+
+        left_tree_inorder = inorder[:root_inorder_index]
+        right_tree_inorder = inorder[root_inorder_index + 1 :]
+
+        return TreeNode(
+            root_val,
+            self.buildTree(preorder, left_tree_inorder),
+            self.buildTree(preorder, right_tree_inorder),
+        )
+
+
+# Overall time complexity: O(N^2)
+# - Finding the root in the inorder list and splitting it takes O(N) time in each call.
+# - There are N nodes, so this operation is repeated N times.
+
+# Overall space complexity: O(N)
+# - The primary space usage is the recursion stack which goes as deep as the height of the tree.
+# - In the worst case (unbalanced tree), this can be O(N).
+# - In the best case (balanced tree), this is O(log N).