lca py3

0235_lowest-common-ancestor-of-a-binary-search-tree/README.md

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+Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
+
+According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): “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:**
+
+![](https://assets.leetcode.com/uploads/2018/12/14/binarysearchtree_improved.png)
+
+    Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
+    Output: 6
+    Explanation: The LCA of nodes 2 and 8 is 6.
+    
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2018/12/14/binarysearchtree_improved.png)
+
+    Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
+    Output: 2
+    Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
+    
+
+**Example 3:**
+
+    Input: root = [2,1], p = 2, q = 1
+    Output: 2
+    
+
+**Constraints:**
+
+*   The number of nodes in the tree is in the range `[2, 105]`.
+*   `-109 <= Node.val <= 109`
+*   All `Node.val` are **unique**.
+*   `p != q`
+*   `p` and `q` will exist in the BST.
+
+https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree

0235_lowest-common-ancestor-of-a-binary-search-tree/python3/solution.py

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+# 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':
+        curr = root
+        
+        while curr:
+            # Both p and q can be found in right subtree so move
+            # curr there
+            if p.val > curr.val and q.val > curr.val:
+                curr = curr.right
+            # Both p and q can be found in left subtree so move
+            # curr there
+            elif p.val < curr.val and q.val < curr.val:
+                curr = curr.left
+            # If above cases fail, that means we are already at a `curr`
+            # where p exists in one side and q exists on another side so 
+            # that must mean `curr` is the LCA
+            else:
+                return curr