LeetCode 426. Convert Binary Search Tree to Sorted Doubly Linked List

看起來很難，但是仔細想一下，實質就是二叉樹的中序遍歷的問題，中序遍歷有遞歸和非遞歸(至少兩種寫法)。

遞歸：

class Solution {
public:
    Node *prev; //實質是指向最后一個元素的指針
    
    Node* treeToDoublyList(Node* root) {
        if (root==NULL) return NULL;
        Node *dummy=new Node(0,NULL,NULL);
        prev = dummy;
        inorder(root);
        
        prev->right = dummy->right;
        dummy->right->left = prev;
        return dummy->right;
    }
    
    void inorder(Node *root){
        if (root==NULL) return;
        inorder(root->left);
        prev->right = root;
        root->left = prev;
        prev = root;
        inorder(root->right);
    }
};

 

非遞歸

class Solution {
public:
    Node *prev; //實質是指向最后一個元素的指針
    
    Node* treeToDoublyList(Node* root) {
        if (root==NULL) return NULL;
        
        stack<Node *> s;
        Node *dummy=new Node(0,NULL,NULL);
        Node *p=root, *prev=dummy;
        
        while (!s.empty() || p!=NULL){
            while (p){
                s.push(p);
                p = p->left;
            }
            p = s.top(), s.pop();
                
            prev->right = p;
            p->left = prev;
            prev = p;
                
            p = p->right;
        }
        
        prev->right = dummy->right;
        dummy->right->left = prev;
        
        return dummy->right;
    }
};

 

class Solution {
public:
    Node *prev; //實質是指向最后一個元素的指針
    
    Node* treeToDoublyList(Node* root) {
        if (root==NULL) return NULL;
        
        Node *dummy=new Node(0,NULL,NULL);
        Node *prev=dummy;
        
        stack<Node *> s({root});
        unordered_map<Node *,int> hash;
        while (!s.empty()){
            Node *cur=s.top(); s.pop();
            if (hash[cur]==0){
                ++hash[cur];
                s.push(cur);
                if (cur->left) s.push(cur->left);
            }else{
                prev->right = cur;
                cur->left = prev;
                prev = cur;
                
                if (cur->right) s.push(cur->right);
            }
        }
        
        prev->right = dummy->right;
        dummy->right->left = prev;
        
        return dummy->right;
    }
};

 

Divide and Conquer

思路和中序遍歷很類似，但是代碼寫起來有一點不一樣。感覺這種方法思路更加清晰。

class Solution {
public:
    Node* treeToDoublyList(Node* root) {
        if (root==NULL) return NULL;
        Node *leftHead=treeToDoublyList(root->left);
        Node *rightHead=treeToDoublyList(root->right);
        root->left = root; root->right = root; // make root a doublylist
        return link(link(leftHead, root),rightHead);
    }
    
    Node *link(Node *head1, Node *head2){
        if (head1==NULL) return head2;
        if (head2==NULL) return head1;
        
        Node *tail1=head1->left;
        Node *tail2=head2->left;
        
        tail1->right = head2;
        head2->left = tail1;
        
        tail2->right = head1;
        head1->left = tail2;
        
        return head1;
    }
};