236. 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).”
Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]
_______3______
/ \
___5__ ___1__
/ \ / \
6 _2 0 8
/ \
7 4
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 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.
Note:
- All of the nodes' values will be unique.
- p and q are different and both values will exist in the binary tree.
Solution
(1) Java
(2) Python
# 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, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
if not root:
return root
if root == p or root == q:
return root
lnode = self.lowestCommonAncestor(root.left, p, q)
rnode = self.lowestCommonAncestor(root.right, p, q)
if lnode and rnode:
return root
return lnode if lnode else rnode
More concise code
def lowestCommonAncestor(self, root, p, q):
if root in (None, p, q): return root
left, right = (self.lowestCommonAncestor(kid, p, q)
for kid in (root.left, root.right))
return root if left and right else left or right
(3) Scala