1650. 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.

Solution:

The last line works because None in python evaluates to False, so doing None or Node will return Node

class Solution(object):
    def lowestCommonAncestor(self, p, q):
        pVals = set()
        def traverse_up(root):
            if root == None or root in pVals:
                return root
            pVals.add(root)
            return traverse_up(root.parent)
            
        return traverse_up(p) or traverse_up(q)

Solution2:

The idea is to store the parents (path) from root to p, and then check q’s path.

class Solution:
    def lowestCommonAncestor(self, p, q):
        path = set()
        while p:
            path.add(p)
            p = p.parent 
        while q not in path:
            q = q.parent 
        return q

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