1676. Lowest Common Ancestor of a Binary Tree IV

Posted on Edited on In

1676. Lowest Common Ancestor of a Binary Tree IV

Description

Difficulty: Medium

Related Topics: Tree, Depth-First Search, Binary Tree

Given the root of a binary tree and an array of TreeNode objects nodes, return the lowest common ancestor (LCA) of all the nodes in nodes. All the nodes will exist in the tree, and all values of the tree’s nodes are unique.

Extending the definition of LCA on Wikipedia: “The lowest common ancestor of n nodes p1, p2, …, pn in a binary tree T is the lowest node that has every pi as a descendant (where we allow a node to be a descendant of itself) for every valid i“. A descendant of a node x is a node y that is on the path from node x to some leaf node.

Example 1:

1
2
3
Input: root = [3,5,1,6,2,0,8,null,null,7,4], nodes = [4,7]
Output: 2
Explanation: The lowest common ancestor of nodes 4 and 7 is node 2.

Example 2:

1
2
3
4
Input: root = [3,5,1,6,2,0,8,null,null,7,4], nodes = [1]
Output: 1
Explanation: The lowest common ancestor of a single node is the node itself.

Example 3:

1
2
3
Input: root = [3,5,1,6,2,0,8,null,null,7,4], nodes = [7,6,2,4]
Output: 5
Explanation: The lowest common ancestor of the nodes 7, 6, 2, and 4 is node 5.

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • -109 <= Node.val <= 109
  • All Node.val are unique.
  • All nodes[i] will exist in the tree.
  • All nodes[i] are distinct.

Hints/Notes

  • Since every node exists, we can just return when we find the first item in the list
  • Provided an array of nodes, but it’s the same as finding the LCA of two nodes

Solution

Language: C++

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, vector<TreeNode*> &nodes) {
        set<TreeNode*> values(nodes.begin(), nodes.end());
        return find(root, values);
    }

    TreeNode* find(TreeNode* root, set<TreeNode*> &values) {
        if (root == nullptr) return root;

        if (values.contains(root)) {
            return root;
        }

        TreeNode* left = find(root->left, values);
        TreeNode* right = find(root->right, values);
        
        if (left && right) return root;

        return left ? left : right;
    }
};