60-Lowest Common Ancestor of a Binary Search Tree

1. Lowest Common Ancestor of a Binary Search Tree My Submissions QuestionEditorial Solution
   Total Accepted: 68335 Total Submissions: 181124 Difficulty: Easy
   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.
本题是BST，是二叉查找树
针对二叉查找树BST的情况

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        assert(p!=NULL&&q!=NULL);
        if(root==NULL) return NULL;  
        if(max(p->val, q->val) < root->val) 
                return lowestCommonAncestor(root->left, p, q);  
        else if(min(p->val, q->val) > root->val) 
                return lowestCommonAncestor(root->right, p, q);  
             else return root;  
    }
};

针对一般的树有以下解决方法：
思路：
判定是否根，是的话返回根，不是
判定p,q的分布，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) {
        assert(p!=NULL&&q!=NULL); //默认p,q不为空，否则为空的不知归为哪个节点
        if(root==NULL)return NULL;
        if(root==p||root==q)return root;
        TreeNode *tmp;
        bool right_p=inTheTree(root->right,p),left_p = inTheTree(root->left,p);
        bool right_q=inTheTree(root->right,q),left_q = inTheTree(root->left,q);
        if((right_p&&left_q)||(right_q&&left_p))return root; //分别在左右子树，返回根
        if(right_p&&right_q)return lowestCommonAncestor(root->right,p,q);//全在右子树，转化为根为右节点的子问题
        if(left_p&&left_q)return lowestCommonAncestor(root->left,p,q);//全在左子树，转化为根为右节点的子问题
        return tmp;
    }
    bool inTheTree(TreeNode* head, TreeNode* p)//判定p是否在树中
    {
        if(head==NULL||p==NULL)return false;
        if(head==p)return true;
        else return inTheTree(head->left,p)||inTheTree(head->right,p);
    }
};