236. Lowest Common Ancestor of a Binary Tree / 235. Lowest Common Ancestor of a Binary Search Tree

Problem Statement

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T 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 will exist in the tree.

Intuition

Approach:

When we encounter p and q we return them 
Using DFS

Whenever we find, left and right not returning null on both
We return that root

Else we return the not null thing

Links

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

Video Links

https://www.youtube.com/watch?v=_-QHfMDde90&ab_channel=takeUforward

Approach 1:

C++

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root == p or root == q or root == nullptr)
            return root;

        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);

        if(left != nullptr and right != nullptr)
            return root;

        return left == nullptr ? right : left;        
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++

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