235. Lowest Common Ancestor of a Binary Search Tree（python+cpp）

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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]

        _______6______
       /              \
    ___2__          ___8__   
   /      \        /      \    
  0       _4       7       9
         /  \
         3   5 

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6. 

Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4 
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of
 itself according to the LCA definition. 

Note:
All of the nodes’ values will be unique.
p and q are different and both values will exist in the BST.

解释：
经典题目，最低公共祖先，用dfs做。
注意这里是BST，所以可以使用BST的特征，如果两个结点都比root小，那么在root的左子树寻找结果，如果两个结点的都比root大，那么在root的右子树上寻找，如果一个结点小于root（或者等于），另一个结点大于root（或者等于），这证明两个结点分别在root的左子树和右子树上，那么root就是他们最低的公共结点，返回root即可。
注意，一个结点可以是自己的祖先。
python代码：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
  def lowestCommonAncestor(self, root, p, q):
      """
      :type root: TreeNode
      :type p: TreeNode
      :type q: TreeNode
      :rtype: TreeNode
      """
      if p.val<root.val and q.val<root.val:
          return self.lowestCommonAncestor(root.left, p, q)
      elif p.val>root.val and q.val>root.val:
          return self.lowestCommonAncestor(root.right, p, q)
      else:
          return root

c++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) {
       
       int val=root->val;
       if(p->val<val&&q->val<val)
           return lowestCommonAncestor(root->left,p,q);
       else if (p->val>val &&q->val>val)
           return lowestCommonAncestor(root->right,p,q);
       return root;
   }
};

总结：
BST可以这么用，普通的二叉树就不行了哟～