LeetCode 1650. Lowest Common Ancestor of a Binary Tree III

Description

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree-iii/

Given two nodes of a binary tree p and q, return their lowest common ancestor (LCA).

Each node will have a reference to its parent node. The definition for Node is below:

class Node {
    public int val;
    public Node left;
    public Node right;
    public Node parent;
}

According to the definition of LCA on Wikipedia: “The lowest common ancestor of two nodes p and q in a tree T is the lowest node 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.val are unique.
  • p != q
  • p and q exist in the tree.

Explanation

The problem interface is different from Lowest Common Ancestor of a Binary Tree. Find the root of the tree, and use the same approach to find the lowest common ancestor between two nodes.

Python Solution

"""
# Definition for a Node.
class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None
        self.parent = None
"""

class Solution(object):
    def lowestCommonAncestor(self, p, q):
        """
        :type node: Node
        :rtype: Node
        """
            
        root = p
        
        while root.parent != None:
            root = root.parent
        
        print (root.val)
        return self.helper(root, p, q)
        
        
    
    def helper(self, root, p, q):
        if not root:
            return None
        
        if root == p or root == q:
            return root
        
        
        left = self.helper(root.left, p, q)
        right = self.helper(root.right, p, q)
        
        if left and right:
            return root
        
        if left:
            return left
        
        if right:
            return right
        
        return None
            
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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