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.

/**
* 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 NULL;
TreeNode* tmp = root;
while (tmp)
{
if (tmp == p || tmp == q)
{//假设相等。则返回该结点
if (tmp == p)
return p;
else
return q;
}
else if ((p->val > tmp->val && q->val < tmp->val) ||
(p->val < tmp->val && q->val > tmp->val))
{//假设一个大于当前结点。一个小于当前结点。则返回当前结点
return tmp;
}
else if (p->val < tmp->val && q->val < tmp->val)
{//假设同一时候小于当前结点,则两个结点都在左子树其中
tmp = tmp->left;
}
else if (p->val > tmp->val && q->val > tmp->val)
{<span style="font-family: Arial, Helvetica, sans-serif;">//假设同一时候大于当前结点,则两个结点都在右子树其中</span>
				tmp = tmp->right;
}
}
return NULL;
}
};

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