1650. Lowest Common Ancestor of a Binary Tree III

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1650. Lowest Common Ancestor of a Binary Tree III

Description

Difficulty: Medium

Related Topics: Hash Table, Tree, Binary Tree

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:

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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:

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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:

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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:

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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.

Hints/Notes

  • 2023/08/30
  • It’s actually find the intersection of 2 lists
  • No solution from leetcode

Solution

Language: C++

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

class Solution {
public:
    Node* lowestCommonAncestor(Node* p, Node * q) {
        Node *h1 = p, *h2 = q;
        while (h1 != h2) {
            if (h1 == nullptr) {
                h1 = q;
            } else {
                h1 = h1->parent;
            }
            if (h2 == nullptr) {
                h2 = p;
            } else {
                h2 = h2->parent;
            }
        }
        return h1;
    }
};