Formatted question description: https://leetcode.ca/all/1650.html

1650. Lowest Common Ancestor of a Binary Tree III

Level

Medium

Description

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:

Image text

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:

Image text

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, 10^5].
  • -10^9 <= Node.val <= 10^9
  • All Node.val are unique.
  • p != q
  • p and q exist in the tree.

Solution

Use a set to store nodes. Start from p and obtain all the nodes from p to the root node using the nodes’ parents. Then start from q and move towards the root. If a node is already in the set, then the node is a common ancestor of p and q. The first common ancestor that is in the set is the lowest common ancestor of p and q.

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node parent;
};
*/

class Solution {
    public Node lowestCommonAncestor(Node p, Node q) {
        Set<Node> set = new HashSet<Node>();
        Node temp = p;
        while (temp != null) {
            set.add(temp);
            temp = temp.parent;
        }
        temp = q;
        while (temp != null) {
            if (set.contains(temp))
                break;
            else
                temp = temp.parent;
        }
        return temp;
    }
}

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