c++实现多叉树树形显示(适合家谱的显示)
发布日期:2022-04-04 06:36:26 浏览次数:21 分类:博客文章

本文共 5756 字,大约阅读时间需要 19 分钟。

多叉树(左兄弟右孩子二叉树)的树形显示

核心代码

 

void positionadd(Multiway_tree*root, int n){	if (!root)return;	Multiway_tree*pcur = root;	stack
s; while (pcur || !s.empty()) { if (pcur) { if (pcur == root->brother)break; pcur->positon += n; s.push(pcur); pcur = pcur->child; } else { pcur = s.top(); s.pop(); pcur = pcur->brother; } }}void Multiway_tree::display(Multiway_tree *root, const int distance){ if (root == NULL) return; root->positon = 0; root->positionmark = " "; int *namenum = new int[sum]; stack
s; Multiway_tree *pcur; pcur = root; int gener = 0; while (pcur || !s.empty()) { if (pcur) { if (pcur->generation > gener)gener = pcur->generation; if (gener) s.push(pcur); pcur = pcur->child; } else { pcur = s.top(); s.pop(); pcur = pcur->brother; } } Multiway_tree **p1 = new Multiway_tree*[gener + 1]; for (int i = 0; i < gener + 1; ++i) p1[i] = 0; pcur = root; while (pcur || !s.empty()) { if (pcur) { if (!(p1[pcur->generation]))p1[pcur->generation] = pcur; else { Multiway_tree*t = p1[pcur->generation]; while (t->next)t = t->next; t->next = pcur; } namenum[pcur->num] = pcur->name.size() ; if (pcur->parent) if (pcur->parent->child == pcur) { Multiway_tree*t = pcur; if (pcur->brother)pcur->positionmark = "/"; else pcur->positionmark = "|"; t->parent->childrennum++; while (t->brother) { t->parent->childrennum++; t->brother->positionmark = "|"; t = t->brother; } if (t != pcur)t->positionmark = "\\"; } s.push(pcur); pcur = pcur->child; } else { pcur = s.top(); s.pop(); pcur = pcur->brother; } } pcur = root; pcur->positon = 0; while (pcur || !s.empty()) { if (pcur) { if (pcur->parent) { int n = 0; if (pcur->parent->child == pcur) { Multiway_tree*t = pcur; pcur->positon = 0; while (t->brother) { t->brother->positon = t->positon + namenum[t->num] + distance; t = t->brother; } int g = t->parent->positon + (namenum[t->parent->num]) / 2 - (t->positon + namenum[t->num]) / 2; if (t->parent->childrennum == 1)pcur->positon = pcur->parent->positon + namenum[pcur->parent->num] / 2 - namenum[pcur->num] / 2; else pcur->positon = g; t = pcur; while (t->brother) { t->brother->positon += g; t = t->brother; } } } s.push(pcur); pcur = pcur->child; } else { pcur = s.top(); s.pop(); pcur = pcur->brother; } } for (int i = gener; i > 0; --i) { for (Multiway_tree*t1 = p1[i]; t1->next; t1 = t1->next) { if (t1->parent->child != t1)continue; for (Multiway_tree*t2 = t1->next; t2; t2 = t2->next) { if (t2->parent == t1->parent)continue; else { Multiway_tree*t1n = t1; while (t1n->brother)t1n = t1n->brother; Multiway_tree*t2n = t2; while (t2n->brother)t2n = t2n->brother; if (! ((t1->positon >= (t2n->positon + namenum[t2n->num])) || (t2->positon >= (t1n->positon + namenum[t1n->num]))) ) { int d = 0; abs(t1->parent->child->positon - t2n->positon - namenum[t2n->num]) > abs(t2->parent->child->positon - t1n->positon - namenum[t1n->num]) ? abs(t2->parent->child->positon - t1n->positon - namenum[t1n->num]) : abs(t1->parent->child->positon - t2n->positon - namenum[t2n->num]); Multiway_tree*t1p = t1, *t2p = t2; while (t1p->parent != t2p->parent) { t1p = t1p->parent; t2p = t2p->parent; } int n = 0; Multiway_tree*temp = t1p; for (temp = t1p, n = 0; temp != t2p && temp; temp = temp->brother) ++n; if (temp != t2p) { for (temp = t2p, n = 0; temp != t1p && temp; temp = temp->brother) ++n; d = abs(t1->parent->child->positon - t2n->positon - namenum[t2n->num]); } else { d = abs(t2->parent->child->positon - t1n->positon - namenum[t1n->num]); } //4??? int averaged = (d + distance) / n + 1; if (t1p->parent->childrennum % 2 == 0) { int childn = 0; temp = t1p->parent->child; while (temp) { childn++; int t = averaged*(childn - t1p->parent->childrennum / 2) - averaged / 2; positionadd(temp, t); temp = temp->brother; } } else { int childn = 0; temp = t1p->parent->child; while (temp) { childn++; int t = averaged*(childn - (t1p->parent->childrennum + 1) / 2); positionadd(temp, t); temp = temp->brother; } } } } } } } int minposition = INT_MAX; pcur = root; while (pcur || !s.empty()) { if (pcur) { if (minposition > pcur->positon) minposition = pcur->positon; s.push(pcur); pcur = pcur->child; } else { pcur = s.top(); s.pop(); pcur = pcur->brother; } } positionadd(root, -1 * minposition + 1); COORD info = GetLargestConsoleWindowSize(GetStdHandle(STD_OUTPUT_HANDLE)); if (root->positon < info.X / 4); positionadd(root, info.X / 4 - root->positon); string *output = new string[(gener + 1) * 3]; for (int i = 0; i < (gener + 1) * 3; ++i) output[i].append(119, ' '); for (int i = 1; i < gener + 1; ++i) { for (Multiway_tree *temp = p1[i]; temp; temp = temp->next) { string t = temp->name ; output[i].replace(temp->positon, namenum[temp->num], t); output[i + gener].replace(temp->positon + namenum[temp->num] / 2, 1, (temp->positionmark)); output[i + gener * 2].replace(temp->positon + namenum[temp->num] / 2, to_string(temp->num).size(), to_string(temp->num)); } } for (int i = 1; i < gener + 1; ++i) { cout << output[i + gener * 2] << endl << output[i] << endl << endl; if (i != gener)cout << output[i + 1 + gener] << endl << endl; } }

  下载地址

 https://download.csdn.net/download/li_haoren/10336965

perorder.txt inorder.txt 分别是对本质的二叉树的前序和中序遍历保存的文件

转载地址:https://www.cnblogs.com/l2017/p/8660089.html 如侵犯您的版权,请留言回复原文章的地址,我们会给您删除此文章,给您带来不便请您谅解!

上一篇:小知识
下一篇:正则表达式匹配器------代码之美

发表评论

最新留言

网站不错 人气很旺了 加油
[***.192.178.218]2024年04月19日 17时51分00秒

关于作者

    喝酒易醉,品茶养心,人生如梦,品茶悟道,何以解忧?唯有杜康!
-- 愿君每日到此一游!

推荐文章