LeetCode *** 235. Lowest Common Ancestor of a Binary Search Tree

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本文探讨了在二叉搜索树中查找两个指定节点的最低公共祖先的问题，并提供了一种有效的算法实现方案。通过对节点间的关系进行判断，算法能够快速定位到目标节点的共同祖先。

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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.

分析：

嗯。。我一开始以为LCA的意思是数字最小。。结果才发现是离得最近的。。那么没啥意思了。。。只要p和q一个在root的左边另外一个在右边，那么root就是所求结果了。。。

代码：

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