[Tree] binary search trees and doubly-linked list

The general idea:

In fact, this binary tree problem mostly with "recursion" to solve. This question is just the beginning I did not find the idea of ​​recursion.

If there are multiple nodes in the sub-tree , you need to find clear in the end what is the root node before and after the . Such as:

May know, this is obviously in accordance with the order tree traversal thing. .

then. . . Analysis for a long time. . . Because preorder binary sort tree originally ordered, so in fact, preorder, just to consider a pre nodes to link the context of it ! !

This code also I studied for a while, pay attention to pre pointer is a reference variable. Code structure is in order traversal of the structure, just to figure out what I currently do, pre pointer should change how.

AC Code:

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        //先判空和一个
        if(pRootOfTree==NULL)
        {
            return NULL;
        }
        if(pRootOfTree->left==NULL && pRootOfTree->right==NULL)
            return pRootOfTree;
        
        TreeNode* pre = NULL;
        convert_recursive(pRootOfTree,pre); //初始的pre值是null
        
        TreeNode* p = pRootOfTree;
        while(p->left)
            p=p->left;
        return p; //结果链表的头结点是树的最左下方的结点
    }
    
    void convert_recursive(TreeNode* t, TreeNode* &pre) //注意pre要加引用
    {
        if(t==NULL) 
            return;
        
        convert_recursive(t->left,pre);
        
        //仿照中序遍历，这里是主要的操作：
        t->left=pre; //注意，这里要理解到！！！要等左子树以同样逻辑得到的pre作为本根节点的在链表中的上一个元素
        if(pre) pre->right = t; //注意是双向链表！
        pre = t; //这步也很关键
        
        convert_recursive(t->right,pre);
    }
};