【LeetCode 235】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__
   /      \        /      \
         _4
         /   \
            

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.

思路：

　　利用BST的性质，若待查找两个结点的数值都小于当前结点则向左边查找（若有一个等于当前结点则当前结点就是其最小的公共结点，结束查找），反之向右边查找（同理），若一个大于一个小于，则当前结点就是其最小的公共结点，结束查找。

C++：

 /**
  * 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 {
 private:
     TreeNode *l, *r, *ret;
 public:

     void rec(TreeNode *root)
     {
         int v = root->val;
         if(l->val <= v && r->val <= v)
         {
             if(v == l->val || v == r->val)
             {
                 ret = root;
                 return ;
             }

             if(root->left != )
                 rec(root->left);
         }
         else if(l->val < v && v < r->val)
         {
             ret = root;
             return ;
         }
         else if(l->val >= v && r->val >= v)
         {
             if(v == l->val || v == r->val)
             {
                 ret = root;
                 return ;
             }

             if(root->right != )
                 rec(root->right);
         }
     }

     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
         if(root == )
             return ;

         if(p->val < q->val){
             l = p;
             r = q;
         }
         else{
             l = q;
             r = p;
         }

         ret = ;

         rec(root);

         return ret;
     }
 };