# [leetcode] Lowest Common Ancestor of a Binary Search Tree

### Lowest Common Ancestor of a Binary Search Tree

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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

```        _______6______
/              \
___2__          ___8__
/      \        /      \
0      _4       7       9
/  \
3   5
```

For example, the lowest common ancestor (LCA) of nodes `2` and `8` is `6`. Another example is LCA of nodes `2` and `4` is `2`, since a node can be a descendant of itself according to the LCA definition.

• if p->val > q->val, swap q and p
• if root == null, reutrn null
• if root->val >= p->val && root->val <= q->val return root; //root is the lowest common ancestor
• if root->val < p->val , recursively find in root->right
• else , recursively find in root->left
```/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
//duplicates in tree?
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(p->val > q->val){
TreeNode * tmp = p;
p = q;
q = tmp;
}
return _lowestCommonAncestor(root, p, q);
}
TreeNode * _lowestCommonAncestor(TreeNode * root, TreeNode * p, TreeNode * q){
if(root == nullptr){
return nullptr;
}
else if(p->val <= root->val && q->val >= root->val){
return root;//find the common ancestor
}
else if(p->val > root->val){
return _lowestCommonAncestor(root->right, p, q);
}
else{
return _lowestCommonAncestor(root->left, p, q);
}

}
};```

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