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: “The lowest common ancestor is defined between two nodes
q as the lowest node in
T that has both
q as descendants (where we allow a node to be a descendant of itself).”
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.
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.
Input: root = [2,1], p = 2, q = 1 Output: 2
- The number of nodes in the tree is in the range
-109 <= Node.val <= 109
p != q
qwill exist in the BST.
If root is either p or q, LCA is root.
If LCA is in either subtree, return the LCA from that subtree.
If p and q are in different subtree, return the root.
# 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': return self.helper(root, p, q) def helper(self, root, p, q): if not root: return root if root == p or root == q: return root left = self.helper(root.left, p, q) right = self.helper(root.right, p, q) if left and right: return root if left: return left if right: return right
- Time Complexity: O(N).
- Space Complexity: O(N).