深信服2020编程题

一.  深信服2020编程第一题. 题目大意是从层次遍历序列得到前序遍历序列...一开始看得有些懵..现在来总结一下.

(因为我自己定义输入输出,所以就没有原题那么严格了.........)

二叉树的序列化

#include 
#include 
using namespace std;

struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

//层次遍历构造树,用队列.
TreeNode* generateTree(queue list) {
	if (list.front() == '#') return NULL;
	//队列里存的都是树节点.
	queue tmp;
	TreeNode* root = new TreeNode(list.front() - '0');
	//tmp存放需要广度优先遍历的节点...
	tmp.push(root);list.pop();
	//list为空说明包括#都没了,才算结束.
	while (!list.empty()) {
		//取出节点.
		TreeNode* node = tmp.front();
		tmp.pop();
		//不等于#则构建左节点.
		if (list.front() != '#') {
			node->left = new TreeNode(list.front() - '0');
			tmp.push(node->left); list.pop();
		}
		else list.pop();
		//接着构建右子树节点.
		if (list.front() != '#') {
			node->right = new TreeNode(list.front() - '0');
			tmp.push(node->right); list.pop();
		}
		else list.pop();
	}
	return root;
}

//通过树递归进行先序遍历.
void preorder(TreeNode* root, vector& res) {
	if (root == NULL) {
		res.push_back(-1);
		return;
	}
	res.push_back(root->val);
	preorder(root->left, res);
	preorder(root->right, res);
}

int main() {
	queue list;
	list.push('0'); list.push('1'); list.push('2'); list.push('3');
	list.push('#'); list.push('#'); list.push('4'); list.push('#');
	list.push('5'); list.push('6'); list.push('#');
	
	TreeNode* root = generateTree(list);
	vector res;
	preorder(root, res);
	for (int i = 0; i < res.size(); i++)
		cout << res[i] << endl;
	cin.get();
	return 0;
}