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
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.
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.
- All of the nodes’ values will be unique.
- p and q are different and both values will exist in the binary tree.
The question becomes finding the smallest subtree that contains both p and q and returning its root. A subtree contains a node is either a node in its left tree, right tree, or its root. Because of the recursive call, we can store the first occurrence of such tree when both p and q exists, and it is the LCA.
# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def lowestCommonAncestor(self, root, p, q): """ :type root: TreeNode :type p: TreeNode :type q: TreeNode :rtype: TreeNode """ self.lca = None self.contains(root, p, q) return self.lca def contains(self, root, p, q): if root is None: return False, False left_p, left_q = self.contains(root.left, p, q) right_p, right_q = self.contains(root.right, p, q) if self.lca is not None: return True, True contains_p = root.val == p.val or left_p or right_p contains_q = root.val == q.val or left_q or right_q if contains_p and contains_q: self.lca = root return contains_p, contains_q