剑指offer 面试题36.二叉搜索树与双向链表

中序递归，一个pre节点记录前一个节点

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    TreeNode* pre=nullptr;
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree==nullptr){
            return nullptr;
        }
        func(pRootOfTree);
        while(pRootOfTree->left){
            pRootOfTree=pRootOfTree->left;
        }
        return pRootOfTree;
    }
    void func(TreeNode* pRootOfTree){
        if(pRootOfTree==nullptr){
            return;
        }
        Convert(pRootOfTree->left);
        if(pre!=nullptr){
            pre->right=pRootOfTree;
            pRootOfTree->left=pre;
        }
        pre=pRootOfTree;
        Convert(pRootOfTree->right);
    }
};

中序迭代，一个pre节点记录前一个节点

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    TreeNode* pre=nullptr;
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree==nullptr){return nullptr;}
        stack<TreeNode*> sta;
        TreeNode* cur=pRootOfTree;
        while(cur or not sta.empty()){
            if(cur){
                sta.push(cur);
                cur=cur->left;
            }
            else{
                cur=sta.top();
                sta.pop();
                if(pre!=nullptr){
                    pre->right=cur;
                    cur->left=pre;
                }
                pre=cur;
                cur=cur->right;
            }
        }
        while(pRootOfTree->left){
            pRootOfTree=pRootOfTree->left;
        }
        return pRootOfTree;
    }
};

递归，func(root)将root为根的子树转化为双向链表，返回长度为2的vector，vec[0]为子树链表的头，vec[1]为尾。

/*
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)
    {
        return func(pRootOfTree)[0];
    }
    vector<TreeNode*> func(TreeNode* root){
        if(root==nullptr){return {nullptr,nullptr};}
        auto vec_le=func(root->left);
        auto vec_ri=func(root->right);
        if(vec_le[1]!=nullptr){
            vec_le[1]->right=root;
            root->left=vec_le[1];
        }
        if(vec_ri[0]!=nullptr){
            vec_ri[0]->left=root;
            root->right=vec_ri[0];
        }
        vector<TreeNode*> res(2,root);
        if(vec_le[0]!=nullptr){
            res[0]=vec_le[0];
        }
        if(vec_ri[1]!=nullptr){
            res[1]=vec_ri[1];
        }
        return res;
    }
};