Leetcode: Lowest Common Ancestor of a Binary Tree

Lowest Common Ancestor of a Binary Tree



Similar Problems:


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

Example 1:

Input: root, p = 5, q = 1
Output: 3
Explanation: The LCA of of nodes 5 and 1 is 3.

Example 2:

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.

Github: code.dennyzhang.com

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        
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