面試題36. 二叉搜索樹與雙向鏈表 面試題37. 序列化二叉樹

 面試題36. 二叉搜索樹與雙向鏈表

遞歸實現

問題：棧溢出：  

[28,-98,67,null,-89,62,null,-97,-25,null,64,null,null,-72,-9,null,null,-88,-41,null,-7,null,-78,-53,null,null,2,-85,-77,-69,-42,-1]

class Node {
public:
	int val;
	Node* left;
	Node* right;

	Node() {}

	Node(int _val) {
		val = _val;
		left = NULL;
		right = NULL;
	}

	Node(int _val, Node* _left, Node* _right) {
		val = _val;
		left = _left;
		right = _right;
	}
};

/*
全局保存頭尾指針
先遞歸左子樹
連接，並更新尾部指針
再遞歸右子樹
*/

Node*head = nullptr, *tail = nullptr;

void inOrder(Node*& root) {
	if (root == nullptr) {
		return;
	}

	inOrder(root->left);

	if (head == nullptr) {
		head = root;
	}

	if (tail != nullptr) {
		tail->right = root;
		root->left = tail;
	}
	tail = root;

	inOrder(root->right);

}

Node* treeToDoublyList(Node* root) {
	if (root == nullptr) {
		return nullptr;
	}
	inOrder(root);

	if (tail&&head) {
		tail->right = head;
		head->left = tail;
	}

	return head;

}


Node* buildBST(Node*t,int num) {
	if (t == nullptr) {
		t = new Node(num);
	}
	else
	{
		if (num < t->val) {
			t->left = buildBST(t->left, num);
		}
		else
		{
			t->right = buildBST(t->right,num);
		}
	}
	return t;
}


Node* creat() {

	Node* t = nullptr;
	int n;
	cin >> n;
	while (n--)
	{
		int num;
		cin >> num;
		t=buildBST(t, num);
	}

	return t;

}


int main()
{
	Node* root=creat();

	treeToDoublyList(root);

	std::cout << "Hello World!\n";
}

 棧

• 將樹的左子樹依次存入stack
• 每次pop出一個Node將其右子樹的左子樹也依次存入stack
• 飯後做連接操作

Node* treeToDoublyList(Node* root) {
	if (root == nullptr) {
		return nullptr;
	}
	Node*cur = root;
	Node*before = nullptr;
	Node*after = nullptr;
	Node*head = nullptr;


	stack<Node*> stac;


	while (cur){
		stac.push(cur);
		head = cur;
		cur = cur->left;
	}
	 
	while (!stac.empty()) {
		cur = stac.top();
		stac.pop();
		Node*tmp = cur->right;
		while (tmp) {
			stac.push(tmp);
			tmp = tmp->left;
		}

		if (stac.empty()) {
			after = head;
		}
		else{
			after = stac.top();
		}

		cur->left = before;
		cur->right = after;
		before = cur;

	}
	head->left = before;

	return head;


}

 

 面試題37. 序列化二叉樹

遞歸

void preOrder(TreeNode* root, string&res) {
	if (root) {
		res += "#,";
		return;
	}

	res += to_string(root->val) + ",";
	preOrder(root->left,res);
	preOrder(root->right,res);
}

// Encodes a tree to a single string.
string serialize(TreeNode* root) {
	string res="";
	if (root == nullptr) {
		return res;
	}

	preOrder(root,res);

	return res;
}
TreeNode* preOrderDecode(int& p, const string& data) {
	if (data[p] == '#') {
		p += 2;
		return nullptr;
	}

	bool is_n = false;
	if (data[p] == '-') {
		is_n = true;
		p++;
	}
	int num = 0;
	while (data[p]!=','){
		num = num * 10 + (data[p] - '0');
		p++;
	}
	if (is_n) num = -num;

	auto root = new TreeNode(num);
	root->left = preOrderDecode(p, data);
	root->right = preOrderDecode(p, data);
    return root;
}
// Decodes your encoded data to tree.
TreeNode* deserialize(string data) {
	if (data.size() <= 0) {
		return nullptr;
	}
	int p = 0;
	return preOrderDecode(p,data);

}
};

 

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