feat: solve DaleStudy#249 with python

Author: EgonD3V

lowest-common-ancestor-of-a-binary-search-tree/EGON.py

@@ -0,0 +1,106 @@
+from collections import deque
+from unittest import TestCase, main
+
+
+# Definition of TreeNode:
+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:
+        return self.solve_bfs(root, p, q)
+
+    """
+    Runtime: 50 ms (Beats 81.68%)
+    Time Complexity:`
+        - bfs를 이용해 ancetor와 nodes를 초기화 하는데 트리의 모든 p와 q를 포함할 때까지 모든 node를 조회하는데, O(n)
+        - p로 부터 부모를 따라 올라가는데, 트리의 모든 node가 편향적으로 연결된 경우 최대 O(n), upper bound
+            - q로 부터 부모를 따라 올라가는데, 단, 위 p의 추적 path와 겹치지 않는 곳만 올라가므로, 총합이 O(n)
+        > O(n) + (O(P) + O(Q)) = O(n) + O(n) ~= O(n) 
+
+    Memory: 21.20 MB (Beats 14.40%)
+    Space Complexity: O(n)
+        - p, q가 트리의 리프노드인 경우 dq의 크기는 O(n), upper bound
+        - p_ancestors와 q_anscestor의 크기는 합쳐서 O(n)
+        > O(n) + O(n) ~= O(n)
+    """
+    def solve_bfs(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:
+        dq = deque([root])
+        ancestor = {root.val: root.val}
+        nodes = {root.val: root}
+        while dq:
+            if p.val in ancestor and q.val in ancestor:
+                break
+
+            curr_node = dq.popleft()
+            if curr_node.left:
+                ancestor[curr_node.left.val] = curr_node.val
+                dq.append(curr_node.left)
+                nodes[curr_node.left.val] = curr_node.left
+            if curr_node.right:
+                ancestor[curr_node.right.val] = curr_node.val
+                dq.append(curr_node.right)
+                nodes[curr_node.right.val] = curr_node.right
+
+        p_val = p.val
+        p_ancestors = set()
+        p_ancestors.add(root.val)
+        while p_val in ancestor and p_val != root.val:
+            p_ancestors.add(p_val)
+            p_ancestors.add(ancestor[p_val])
+            p_val = ancestor[p_val]
+
+        q_val = q.val
+        q_ancestors = set()
+        q_ancestors.add(q_val)
+        while q_val in ancestor and q_val not in p_ancestors:
+            q_ancestors.add(q_val)
+            q_ancestors.add(ancestor[q_val])
+            q_val = ancestor[q_val]
+
+        common_ancestor = p_ancestors & q_ancestors
+        if common_ancestor:
+            return nodes[common_ancestor.pop()]
+        else:
+            return root
+
+
+class _LeetCodeTestCases(TestCase):
+
+    def test_1(self):
+        root = TreeNode(5)
+        node1 = TreeNode(3)
+        node2 = TreeNode(6)
+        node3 = TreeNode(2)
+        node4 = TreeNode(4)
+        node5 = TreeNode(1)
+
+        root.left = node1
+        root.right = node2
+        node1.left = node3
+        node1.right = node4
+        node3.left = node5
+
+        p = node5
+        q = node1
+        output = root
+        self.assertEqual(Solution().lowestCommonAncestor(root, p, q), output)
+
+    def test_2(self):
+        root = TreeNode(2)
+        node1 = TreeNode(1)
+
+        root.left = node1
+
+        p = TreeNode(2)
+        q = TreeNode(1)
+        output = root
+        self.assertEqual(Solution().lowestCommonAncestor(root, p, q), output)
+
+
+if __name__ == '__main__':
+    main()