leetcode

commit 618aeaf7949007df444f6f73de0c0a2c85fcd097
parent 4359a7cf6ee7bd279e92d1a1422660a0456e5053
Author: Andrew Laack <andrew@laack.co>
Date:   Fri, 27 Jun 2025 18:24:21 -0500

Completed 2 more problems

Diffstat:
A
lowest-common-ancestor-of-a-binary-search-tree/lowest.py
 | 
27
+++++++++++++++++++++++++++
A
same-tree/same-tree.py
 | 
31
+++++++++++++++++++++++++++++++

2 files changed, 58 insertions(+), 0 deletions(-)
diff --git a/lowest-common-ancestor-of-a-binary-search-tree/lowest.py b/lowest-common-ancestor-of-a-binary-search-tree/lowest.py
@@ -0,0 +1,27 @@
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
+        if p.val > q.val:
+            temp = p
+            p = q
+            q = temp
+        
+        return dfs(root, p, q)
+
+# Assume input of p <= q
+def dfs(current, p, q):
+    assert not current is None
+
+    if p.val <= current.val and q.val >= current.val:
+        return current
+    
+    if current.val > p.val:
+        return dfs(current.left, p, q)
+    else:
+        return dfs(current.right, p, q)
diff --git a/same-tree/same-tree.py b/same-tree/same-tree.py
@@ -0,0 +1,31 @@
+# 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 isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
+        right = p
+        left = q
+        return dfs(left,right)
+    
+
+def dfs(left, right):
+
+    # base case
+    none = 0
+    if right is None:
+        none += 1
+    if left is None:
+        none += 1    
+    if none == 1:
+        return False
+    if none == 2:
+        return True
+    
+    left_id = dfs(left.left, right.left)
+    right_id = dfs(left.right, right.right)
+    current_id = left.val == right.val
+
+    return bool(min(left_id, right_id, current_id))