牛客OJ：二叉搜索树转双向列表

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采用中序遍历：

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    void dfs(vector<TreeNode*>& v,TreeNode* p){
        if(p == NULL) return ;
        dfs(v,p->left);
        v.push_back(p);
        dfs(v,p->right);
    }
    void doit(vector<TreeNode*> v){
        int l = v.size();
        v[0]->left = NULL;
        for(int i=1;i<l;i++){
            v[i-1]->right = v[i];
            v[i]->left = v[i-1];
        }
        v[l-1]->right = NULL;
    }
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree == NULL) return NULL;
        vector<TreeNode*> v;
        dfs(v,pRootOfTree);
        doit(v);
        return v[0];
    }
};