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
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 = [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.
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.
Input: root = [1,2], p = 1, q = 2 Output: 1
- The number of nodes in the tree is in the range
-109 <= Node.val <= 109
p != q
qwill exist in the tree.
If p and q in the left and right subtrees, return root.
If in one side of the subtree, find the LCA in that subtree.
If the root is p or q or is null, return 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.lca_helper(root, p, q) def lca_helper(self, root, p, q): if not root: return None if p == root or q == root: return root left_lca = self.lca_helper(root.left, p, q) right_lca = self.lca_helper(root.right, p, q) if left_lca and right_lca: return root if left_lca: return left_lca if right_lca: return right_lca return None
- Time Complexity: O(N).
- Space Complexity: O(1).
One Thought to “LeetCode 236. Lowest Common Ancestor of a Binary Tree”
Are you sure the space complexity is O(1) if you use recursion?