235 Lowest Common Ancestor of a Binary Search Tree 二叉搜索树的最近公共祖先

给定一棵二叉搜索树, 找到该树中两个指定节点的最近公共祖先。

详见：https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/

Java实现：

方法一：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root==null){
            return null;
        }
        if(root.val>Math.max(p.val,q.val)){
            return lowestCommonAncestor(root.left,p,q);
        }
        if(root.val<Math.min(p.val,q.val)){
            return lowestCommonAncestor(root.right,p,q);
        }
        return root;
    }
}

 方法二：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root==null||root==p||root==q){
            return root;
        }
        TreeNode left=lowestCommonAncestor(root.left,p,q);
        TreeNode right=lowestCommonAncestor(root.right,p,q);
        if(left!=null&&right!=null){
            return root;
        }
        return left!=null?left:right;
    }
}

 C++实现:

方法一：

/**
 * 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==nullptr)
        {
            return root;
        }
        if(root->val>max(p->val,q->val))
        {
            return lowestCommonAncestor(root->left,p,q);
        }
        if(root->val<min(p->val,q->val))
        {
            return lowestCommonAncestor(root->right,p,q);
        }
        return root;
    }
};

方法二：

/**
 * 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==nullptr||root==p||root==q)
        {
            return root;
        }
        TreeNode *left=lowestCommonAncestor(root->left,p,q);
        TreeNode *right=lowestCommonAncestor(root->right,p,q);
        if(left&&right)
        {
            return root;
        }
        return left?left:right;
    }
};

 参考：http://www.cnblogs.com/grandyang/p/4640572.html