feat(work): add solution for Lowest Common Ancestor of a Binary Tree

work/1/Medium/Tree/236.lowest-common-ancestor-of-a-binary-tree.py

@@ -0,0 +1,76 @@
+"""
+236. Lowest Common Ancestor of a Binary Tree
+Difficulty: Medium
+https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/
+
+──────────────────────────────────────────────────
+
+Given a binary tree, find the lowest common ancestor (LCA) of two
+given nodes in the tree.
+
+According to the definition of LCA on Wikipedia: “The lowest
+common ancestor is defined between two nodes p and q as the lowest
+node in T that has both p and q as descendants (where we allow a node
+to be a descendant of itself).”
+
+
+
+Example 1:
+
+Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
+Output: 3
+Explanation: The LCA of nodes 5 and 1 is 3.
+
+Example 2:
+
+Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
+Output: 5
+Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a
+descendant of itself according to the LCA definition.
+
+Example 3:
+
+Input: root = [1,2], p = 1, q = 2
+Output: 1
+
+
+
+Constraints:
+
+        • The number of nodes in the tree is in the range [2, 10^5].
+
+        • -10^9 <= Node.val <= 10^9
+
+        • All Node.val are unique.
+
+        • p != q
+
+        • p and q will exist in the tree.
+"""
+
+# 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 not root:
+            return None
+
+        if root == p or root == q:
+            return root
+
+        l = self.lowestCommonAncestor(root.left, p, q)
+        r = self.lowestCommonAncestor(root.right, p, q)
+
+        if l and r:
+            return root
+        else:
+            return l or r