Lowest Common Ancestor of a Binary Tree
- Review: Classic Code Problems
- Lowest Common Ancestor of a Binary Search Tree
- Tag: #classic, #manydetails, #recursive, #inspiring
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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”
Given the following binary search tree: root = [3,5,1,6,2,0,8,null,null,7,4]
_______3______ / \ ___5__ ___1__ / \ / \ 6 _2 0 8 / \ 7 4
Input: root, p = 5, q = 1 Output: 3 Explanation: The LCA of of nodes 5 and 1 is 3.
Input: root, 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.
Credits To: leetcode.com
Leave me comments, if you have better ways to solve.
## Blog link: https://code.dennyzhang.com/lowest-common-ancestor-of-a-binary-tree ## Basic Ideas: recursive ## ## Notice: ## Here we assume p, q will exists in the tree ## ## Complexity: ? # 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 """ if root is None or root == p or root == q: return root left = self.lowestCommonAncestor(root.left, p, q) right = self.lowestCommonAncestor(root.right, p, q) if left and right: return root return left if left else right