235. Lowest Common Ancestor of a Binary Search Tree.cpp

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

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).”

Given binary search tree:  root = [6,2,8,0,4,7,9,null,null,3,5]




Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.
Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.


Note:

All of the nodes' values will be unique.
p and q are different and both values will exist in the BST.


Solution one using recursive
Solution two is iterative
*/

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if ( !root ) return nullptr;
        if ( root->val > p->val && root->val > q->val )
            return lowestCommonAncestor( root->left, p, q );
        if ( root->val < p->val && root->val < q->val )
            return lowestCommonAncestor( root->right, p, q );
        return root;
    }
};

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        bool qV, pV;
        while ( 1 )
        {
            if ( root->val == p->val || root->val == q->val ) return root;
            if ( p->val < root->val )
                    pV = 1;
            else
                    pV = 0;
            if ( q->val < root->val )
                    qV = 1;
            else
                    qV = 0;
            if ( pV != qV ) return root;
            else if ( !pV ) root = root->right;
            else if ( pV ) root = root->left;
        }
        return root;
    }
};