[Leetocde]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 allowa 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.

/**
* 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:
/*algorithm
1)p and q is in separated child tree, root is their LCA
2)p or q is root, return root
3)recursive call lowestCommonAncestor
further thoughts:
1)p or q is not in the root tree, how about this?
2)how about to handle this if either of root,p,q is NULL
3)how about p and q is same node? return its parent, this seems reasonable

*/

//parent
TreeNode* parent(TreeNode* root,TreeNode* p){

if(p == root)return NULL;
if(root->right == p ||
root->left == p)return root
parent(root->left,p);
parent(root->right,p);
}
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
//handle case 2 root,p,or q is null
if(!root || !p || !q)return NULL;
//handle 3 case
if(p == q)return parent(p);

//handle 1 issue, p or q is not in the tree
TreeNode* first,*last;
first = root;
while(first->left)first = first->left;
last = root;
while(last->right)last = last->right;
if(first->val > p->val || first->val > q->val ||
last->val < p->val || last->val < q->val)//not in the tree
return NULL;
//just consider normal condition
if(root->val > p->val && root->val > q->val) //p and q in left child
return lowestCommonAncestor(root->left,p,q);
if(root->val < p->val && root->val < q->val) //p and q in right
return lowestCommonAncestor(root->right,p,q);
//otherwise , p and q one is in left, one is on right
return root;
}
}