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, 105]. -
-109 <= Node.val <= 109 - All
Node.valare unique. -
p != q -
pandqwill exist in the tree.
SOLUTION:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# class Solution:
# def isInTree(self, root, node):
# if root:
# if root.val == node.val:
# return True
# if self.isInTree(root.left, node) or self.isInTree(root.right, node):
# return True
# return False
# def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
# if root:
# if root.val == p.val or root.val == q.val:
# return root
# pInLeft = self.isInTree(root.left, p)
# qInLeft = self.isInTree(root.left, q)
# if pInLeft and qInLeft:
# return self.lowestCommonAncestor(root.left, p, q)
# if not pInLeft and not qInLeft:
# return self.lowestCommonAncestor(root.right, p, q)
# if pInLeft ^ qInLeft:
# return root
class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
nodepaths = [None, None]
paths = [[root]]
while len(paths) > 0:
curr = paths.pop()
if curr[-1].val == p.val:
nodepaths[0] = curr
if curr[-1].val == q.val:
nodepaths[1] = curr
if nodepaths[0] and nodepaths[1]:
break
if curr[-1].left:
paths.append(curr + [curr[-1].left])
if curr[-1].right:
paths.append(curr + [curr[-1].right])
i = 0
k = min(len(nodepaths[0]), len(nodepaths[1]))
while nodepaths[0][i].val == nodepaths[1][i].val:
i += 1
if i >= k:
break
return nodepaths[0][i - 1]
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