Leetcode题目：Lowest Common Ancestor of a Binary Search Tree

题目：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.

题目解答：这个题目非常简单，就是求在二叉搜索树中，两个子节点的最近公共祖先。

首先说明一下，在二叉排序树中，是没有取值相同的元素的。另外，它还具有以下特点：

二叉排序树或者是一棵空树，或者是具有下列性质的二叉树：

（1）若左子树不空，则左子树上所有结点的值均小于它的根结点的值；

（2）若右子树不空，则右子树上所有结点的值均大于它的根结点的值；

（3）左、右子树也分别为二叉排序树；

代码如下：

/**
* 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 == NULL) || (p == NULL) || (q == NULL))
            return NULL;
        TreeNode * curNode = root;
        if(p -> val > q -> val)
        {
            while(curNode != NULL)
            {
                if(curNode -> val > p -> val)
                {
                    curNode = curNode -> left;
                }
                else if(curNode -> val < q -> val)
                {
                    curNode =curNode -> right;
                }
                else
                    return curNode;
            }
        }
        else if(p -> val < q -> val)
        {
            while(curNode != NULL)
            {
                if(curNode -> val > q -> val)
                {
                    curNode = curNode -> left;
                }
                else if(curNode -> val < p -> val)
                {
                    curNode =curNode -> right;
                }
                else
                    return curNode;
            }
        }
        return NULL;
    }
};