235. Lowest Common Ancestor of a Binary Search Tree

O(N) Time recursive solution

• we traverse the tree and find p and q;
• if one of child node is null return another
• else both are not null return the root(curr node), that means left and right are p and q.

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

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

        if(left == NULL) return right;
        else if(right == NULL) return left;
        else return root; // both are not null
    }
};

O(N) Time iterative solution

• Self Explanatory.

Code

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        TreeNode* cur = root;
        while (true) {
            if (p -> val < cur -> val && q -> val < cur -> val) {
                cur = cur -> left;
            } else if (p -> val > cur -> val && q -> val > cur -> val) {
                cur = cur -> right;
            } else {
                break;
            }
        }
        return cur;
    }
};