蓝桥杯横向二叉树解题报告_蓝桥杯考数据结构么-优快云博客

本文介绍了一种使用C++实现的树形数据结构可视化方法。通过递归算法和字符串操作，该方法能够清晰地展示出树节点之间的关系，并且通过预处理字符串来表示节点间的连接。适用于二叉搜索树等树形结构的可视化输出。

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树形数据结构，输出的时候采用prestr保存数字之前的.和|

#include <iostream>
#include <string>
#include <fstream>
#include <cmath>
#include <cstring>

using namespace std;

struct BTNode{
	int data, len, dis;
	BTNode *lchild, *rchild;
}*root;

char prestr[1000],back[1000];

void Create(BTNode *bt, int value, BTNode *father){
	if (bt == NULL){
		bt = new BTNode;
		bt->data = value;
		bt->dis = father->dis + father->len;
		bt->len = log10(value) + 4.00001;
		bt->lchild = bt->rchild = NULL;
		if (value > father->data)
			father->rchild = bt;
		else
			father->lchild = bt;
	}
	else if (bt->data < value)
		Create(bt->rchild, value, bt);
	else
		Create(bt->lchild, value, bt);
}

void PostOrder(BTNode* bt,bool left,bool right,int pos){
	if (bt != NULL){
		if (bt->rchild){
			if (right){
				//strncpy(back, prestr, 1000);
				//memset(prestr, '.', bt->dis + bt->len);
				prestr[pos] = '.';
				prestr[bt->dis + bt->len + 1] = '|';
				PostOrder(bt->rchild, false, true, bt->dis + bt->len + 1);
				prestr[pos] = '|';
				prestr[bt->dis + bt->len + 1] = '.';
				//strncpy(prestr, back, 1000);
			}
			else
			{
				prestr[bt->dis + bt->len + 1] = '|';
				PostOrder(bt->rchild, false, true, bt->dis + bt->len + 1);
				prestr[bt->dis + bt->len + 1] = '.';
			}
			
		}
		if (bt != root){
			int n = bt->dis;
			for(int i=1;i<=n;i++){
				cout << prestr[i];
			}
			cout << "|-";
		}
		cout << bt->data;
		if (bt->lchild||bt->rchild)
			cout<<"-|" ;
		cout << endl;
		if (bt->lchild){
			if (left){
				//strncpy(back, prestr, 1000);
				//memset(prestr, '.', bt->dis + bt->len);
				prestr[pos] = '.';
				prestr[bt->dis + bt->len + 1] = '|';
				PostOrder(bt->lchild, true, false, bt->dis + bt->len + 1);
				prestr[bt->dis + bt->len + 1] = '.';
				prestr[pos] = '|';
				//strncpy(prestr, back, 1000);
			}
			else
			{
				prestr[bt->dis + bt->len + 1] = '|';
				PostOrder(bt->lchild, true, false, bt->dis + bt->len + 1);
				prestr[bt->dis + bt->len + 1] = '.';
			}
			
		}
	}
}

int main()
{
	//ifstream cin("D:\\input.txt");
	memset(prestr, '.', 1000);
	root = new BTNode;
	cin >> root->data;
	root->len = log10(root->data) + 2.00001;
	root->dis = 0;
	root->lchild = root->rchild = NULL;

	int value;
	while (cin >> value){
		Create(root, value, root);
	}

	PostOrder(root,false,false,0);
	//system("PAUSE");
	return 0;
}