C如何将“二叉树”绘制到控制台上

可以使用哪些算法在控制台中绘制二叉树？该树用C实现。例如，编号为：2 3 4 5 8的BST在控制台中显示为：

在Ascii中签出打印二叉树

来自下面的@AnyOneElse Pastbin：

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!Code originally from /http://www.openasthra.com/c-tidbits/printing-binary-trees-in-ascii/
!!! Just saved it, cause the website is down.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Printing Binary Trees in Ascii

Here we are not going to discuss what binary trees are (please refer this, if you are looking for binary search trees), or their operations but printing them in ascii.

The below routine prints tree in ascii for a given Tree representation which contains list of nodes, and node structure is this

    struct Tree 
    {
      Tree * left, * right;
      int element;
    };

This pic illustrates what the below routine does on canvas..
ascii tree

Here is the printing routine..

    b5855d39a6b8a2735ddcaa04a404c125001 

Auxiliary routines..

    //This function prints the given level of the given tree, assuming
    //that the node has the given x cordinate.
    void print_level(asciinode *node, int x, int level) 
    {
      int i, isleft;
      if (node == NULL) return;
      isleft = (node->parent_dir == -1);
      if (level == 0) 
      {
        for (i=0; i<(x-print_next-((node->lablen-isleft)/2)); i++) 
        {
          printf(" ");
        }
        print_next += i;
        printf("%s", node->label);
        print_next += node->lablen;
      } 
      else if (node->edge_length >= level) 
      {
        if (node->left != NULL) 
        {
          for (i=0; i<(x-print_next-(level)); i++) 
          {
            printf(" ");
          }
          print_next += i;
          printf("/");
          print_next++;
        }
        if (node->right != NULL) 
        {
          for (i=0; i<(x-print_next+(level)); i++) 
          {
            printf(" ");
          }
          print_next += i;
          printf("\\");
          print_next++;
        }
      } 
      else 
      {
        print_level(node->left, 
                    x-node->edge_length-1, 
                    level-node->edge_length-1);
        print_level(node->right, 
                    x+node->edge_length+1, 
                    level-node->edge_length-1);
      }
    }


    //This function fills in the edge_length and 
    //height fields of the specified tree
    void compute_edge_lengths(asciinode *node) 
    {
      int h, hmin, i, delta;
      if (node == NULL) return;
      compute_edge_lengths(node->left);
      compute_edge_lengths(node->right);

      /* first fill in the edge_length of node */
      if (node->right == NULL && node->left == NULL) 
      {
        node->edge_length = 0;
      } 
      else 
      {
        if (node->left != NULL) 
        {
          for (i=0; i<node->left->height && i < MAX_HEIGHT; i++) 
          {
            rprofile[i] = -INFINITY;
          }
          compute_rprofile(node->left, 0, 0);
          hmin = node->left->height;
        } 
        else 
        {
          hmin = 0;
        }
        if (node->right != NULL) 
        {
          for (i=0; i<node->right->height && i < MAX_HEIGHT; i++) 
          {
            lprofile[i] = INFINITY;
          }
          compute_lprofile(node->right, 0, 0);
          hmin = MIN(node->right->height, hmin);
        } 
        else 
        {
          hmin = 0;
        }
        delta = 4;
        for (i=0; i<hmin; i++) 
        {
          delta = MAX(delta, gap + 1 + rprofile[i] - lprofile[i]);
        }

        //If the node has two children of height 1, then we allow the
        //two leaves to be within 1, instead of 2 
        if (((node->left != NULL && node->left->height == 1) ||
              (node->right != NULL && node->right->height == 1))&&delta>4) 
        {
          delta--;
        }

        node->edge_length = ((delta+1)/2) - 1;
      }

      //now fill in the height of node
      h = 1;
      if (node->left != NULL) 
      {
        h = MAX(node->left->height + node->edge_length + 1, h);
      }
      if (node->right != NULL) 
      {
        h = MAX(node->right->height + node->edge_length + 1, h);
      }
      node->height = h;
    }

    asciinode * build_ascii_tree_recursive(Tree * t) 
    {
      asciinode * node;

      if (t == NULL) return NULL;

      node = malloc(sizeof(asciinode));
      node->left = build_ascii_tree_recursive(t->left);
      node->right = build_ascii_tree_recursive(t->right);

      if (node->left != NULL) 
      {
        node->left->parent_dir = -1;
      }

      if (node->right != NULL) 
      {
        node->right->parent_dir = 1;
      }

      sprintf(node->label, "%d", t->element);
      node->lablen = strlen(node->label);

      return node;
    }


    //Copy the tree into the ascii node structre
    asciinode * build_ascii_tree(Tree * t) 
    {
      asciinode *node;
      if (t == NULL) return NULL;
      node = build_ascii_tree_recursive(t);
      node->parent_dir = 0;
      return node;
    }

    //Free all the nodes of the given tree
    void free_ascii_tree(asciinode *node) 
    {
      if (node == NULL) return;
      free_ascii_tree(node->left);
      free_ascii_tree(node->right);
      free(node);
    }

    //The following function fills in the lprofile array for the given tree.
    //It assumes that the center of the label of the root of this tree
    //is located at a position (x,y).  It assumes that the edge_length
    //fields have been computed for this tree.
    void compute_lprofile(asciinode *node, int x, int y) 
    {
      int i, isleft;
      if (node == NULL) return;
      isleft = (node->parent_dir == -1);
      lprofile[y] = MIN(lprofile[y], x-((node->lablen-isleft)/2));
      if (node->left != NULL) 
      {
        for (i=1; i <= node->edge_length && y+i < MAX_HEIGHT; i++) 
        {
          lprofile[y+i] = MIN(lprofile[y+i], x-i);
        }
      }
      compute_lprofile(node->left, x-node->edge_length-1, y+node->edge_length+1);
      compute_lprofile(node->right, x+node->edge_length+1, y+node->edge_length+1);
    }

    void compute_rprofile(asciinode *node, int x, int y) 
    {
      int i, notleft;
      if (node == NULL) return;
      notleft = (node->parent_dir != -1);
      rprofile[y] = MAX(rprofile[y], x+((node->lablen-notleft)/2));
      if (node->right != NULL) 
      {
        for (i=1; i <= node->edge_length && y+i < MAX_HEIGHT; i++) 
        {
          rprofile[y+i] = MAX(rprofile[y+i], x+i);
        }
      }
      compute_rprofile(node->left, x-node->edge_length-1, y+node->edge_length+1);
      compute_rprofile(node->right, x+node->edge_length+1, y+node->edge_length+1);
    }

Here is the asciii tree structure…

    struct asciinode_struct
    {
      asciinode * left, * right;

      //length of the edge from this node to its children
      int edge_length; 

      int height;      

      int lablen;

      //-1=I am left, 0=I am root, 1=right   
      int parent_dir;   

      //max supported unit32 in dec, 10 digits max
      char label[11];  
    };

输出：

        2
       / \
      /   \
     /     \
    1       3
   / \     / \
  0   7   9   1
 /   / \     / \
2   1   0   8   8
       /
      7

码：

int _print_t(tnode *tree, int is_left, int offset, int depth, char s[20][255])
{
    char b[20];
    int width = 5;

    if (!tree) return 0;

    sprintf(b, "(%03d)", tree->val);

    int left  = _print_t(tree->left,  1, offset,                depth + 1, s);
    int right = _print_t(tree->right, 0, offset + left + width, depth + 1, s);

#ifdef COMPACT
    for (int i = 0; i < width; i++)
        s[depth][offset + left + i] = b[i];

    if (depth && is_left) {

        for (int i = 0; i < width + right; i++)
            s[depth - 1][offset + left + width/2 + i] = '-';

        s[depth - 1][offset + left + width/2] = '.';

    } else if (depth && !is_left) {

        for (int i = 0; i < left + width; i++)
            s[depth - 1][offset - width/2 + i] = '-';

        s[depth - 1][offset + left + width/2] = '.';
    }
#else
    for (int i = 0; i < width; i++)
        s[2 * depth][offset + left + i] = b[i];

    if (depth && is_left) {

        for (int i = 0; i < width + right; i++)
            s[2 * depth - 1][offset + left + width/2 + i] = '-';

        s[2 * depth - 1][offset + left + width/2] = '+';
        s[2 * depth - 1][offset + left + width + right + width/2] = '+';

    } else if (depth && !is_left) {

        for (int i = 0; i < left + width; i++)
            s[2 * depth - 1][offset - width/2 + i] = '-';

        s[2 * depth - 1][offset + left + width/2] = '+';
        s[2 * depth - 1][offset - width/2 - 1] = '+';
    }
#endif

    return left + width + right;
}

void print_t(tnode *tree)
{
    char s[20][255];
    for (int i = 0; i < 20; i++)
        sprintf(s[i], "%80s", " ");

    _print_t(tree, 0, 0, 0, s);

    for (int i = 0; i < 20; i++)
        printf("%s\n", s[i]);
}

输出：

                           .----------------------(006)-------.                 
                      .--(001)-------.                   .--(008)--.            
                 .--(-02)       .--(003)-------.       (007)     (009)          
       .-------(-06)          (002)       .--(005)                              
  .--(-08)--.                           (004)                                   
(-09)     (-07)                     

要么

                                                  (006)                         
                           +------------------------+---------+                 
                         (001)                              (008)               
                      +----+---------+                   +----+----+            
                    (-02)          (003)               (007)     (009)          
                 +----+         +----+---------+                                
               (-06)          (002)          (005)                              
       +---------+                        +----+                                
     (-08)                              (004)                                   
  +----+----+                                                                   
(-09)     (-07)                                                       

一些提示：在相同深度（例如，您的示例中为2和4或3和8）的节点之间的间距是深度的函数。

从最左边的节点到最右边的节点，每个打印行都包含具有相同深度的所有节点。

因此，您需要一种方法，例如，根据节点的深度，按照其最左边的顺序将它们排列成行阵列。

从根节点开始，广度优先搜索将按深度和最左侧的顺序访问节点。

节点之间的间距可以通过以下方法找到：找到树的最大高度，对最深的节点使用某个恒定的宽度，对每一个较小的深度使用该宽度加倍，以便任何深度的宽度=（1 + maxdepth-currentdepth）* deepestwidth 。

该数字为您提供任何特定深度的每个节点的打印“水平宽度”。

左节点水平位于其父级宽度的左半部分，右节点位于右半部分。您将为没有父节点的任何节点插入虚拟垫片。一种更简单的方法是确保所有叶子与最深节点处于相同的深度，其值为空白。显然，您还必须补偿值的宽度，也许是通过使最大深度的宽度至少与最大值节点的打印（假定为十进制表示）宽度相同。

当在数组中实现一棵树时，还有另外一个方法：

#include <stdio.h>
#include <math.h>


#define PARENT(i) ((i-1) / 2)
#define NUM_NODES 15
#define LINE_WIDTH 70

int main() {
    int tree[NUM_NODES]={0,1,2,3,4,5,6,7,8,9,1,2,3,4,5};
    int print_pos[NUM_NODES];
    int i, j, k, pos, x=1, level=0;

    print_pos[0] = 0;
    for(i=0,j=1; i<NUM_NODES; i++,j++) {
        pos = print_pos[PARENT(i)] + (i%2?-1:1)*(LINE_WIDTH/(pow(2,level+1))+1);

        for (k=0; k<pos-x; k++) printf("%c",i==0||i%2?' ':'-');
        printf("%d",tree[i]);

        print_pos[i] = x = pos+1;
        if (j==pow(2,level)) {
            printf("\n");
            level++;
            x = 1;
            j = 0;
        }
    }
    return 0;
}

输出：

                                   0
                  1-----------------------------------2
          3-----------------4                 5-----------------6
      7---------8       9---------1       2---------3       4---------5

我在c ++中有这个小解决方案-它可以很容易地转换为c。

我的解决方案确实需要一个补充数据结构来将当前节点的深度存储在树中（这是因为如果您使用的是不完整的树，则给定子树的深度可能与整个树中的深度不一致）。

#include <iostream>
#include <utility>
#include <algorithm>
#include <list>

namespace tree {

template<typename T>
struct node
{
  T data;
  node* l;
  node* r;
  node(T&& data_ = T()) : data(std::move(data_)), l(0), r(0) {}
};

template<typename T>
int max_depth(node<T>* n)
{
  if (!n) return 0;
  return 1 + std::max(max_depth(n->l), max_depth(n->r));
}

template<typename T>
void prt(node<T>* n)
{
  struct node_depth
  {
    node<T>* n;
    int lvl;
    node_depth(node<T>* n_, int lvl_) : n(n_), lvl(lvl_) {}
  };

  int depth = max_depth(n);

  char buf[1024];
  int last_lvl = 0;
  int offset = (1 << depth) - 1;

  // using a queue means we perform a breadth first iteration through the tree
  std::list<node_depth> q;

  q.push_back(node_depth(n, last_lvl));
  while (q.size())
  {
    const node_depth& nd = *q.begin();

    // moving to a new level in the tree, output a new line and calculate new offset
    if (last_lvl != nd.lvl)
    {
      std::cout << "\n";

      last_lvl = nd.lvl;
      offset = (1 << (depth - nd.lvl)) - 1;
    }

    // output <offset><data><offset>
    if (nd.n)
      sprintf(buf, " %*s%d%*s", offset, " ", nd.n->data, offset, " ");
    else
      sprintf(buf, " %*s", offset << 1, " ");
    std::cout << buf;

    if (nd.n)
    {
      q.push_back(node_depth(nd.n->l, last_lvl + 1));
      q.push_back(node_depth(nd.n->r, last_lvl + 1));
    }

    q.pop_front();
  }
  std::cout << "\n";
}

}

int main()
{
  typedef tree::node<int> node;
  node* head = new node();
  head->l    = new node(1);
  head->r    = new node(2);
  head->l->l = new node(3);
  head->l->r = new node(4);
  head->r->l = new node(5);
  head->r->r = new node(6);

  tree::prt(head);

  return 0;
}

它打印出以下内容：

        0                                                                                                
    1       2                                                                                            
  3   4   5   6                                                                                          

查看Linux中pstree命令的输出。它不会以您想要的确切形式生成输出，但是恕我直言，这种方式更具可读性。

我第二点建议。我最近不得不执行此操作以打印Windows进程的VAD树，并且使用了DOT语言（只需从二进制树遍历函数中打印出节点）即可：

http://en.wikipedia.org/wiki/DOT_language

例如，您的DOT文件将包含：

有向图graphname {
     5-> 3;
     5-> 8;
     3-> 4;
     3-> 2;
}

您可以使用dotty.exe生成图形，也可以使用dot.exe将其转换为PNG。

一个非常简单的C ++解决方案在水平方向上打印树：

5
  1
    5
  9
    7
    14

代码（Node::print()功能很重要）：

#include<iostream>

using namespace std;

class Tree;

class Node{
public:
    Node(int val): _val(val){}
    int val(){ return _val; }
    void add(Node *temp)
    {
        if (temp->val() > _val)
        {
            if (_rchild)
                _rchild->add(temp);
            else
            {
                _rchild = temp;
            }
        }
        else
        {
            if (_lchild)
                _lchild->add(temp);
            else
            {
                _lchild = temp;
            }
        }
    }
    void print()
    {
        for (int ix = 0; ix < _level; ++ix) cout << ' ';
        cout << _val << endl;
        ++_level;
        if (_lchild)
        {
            _lchild->print();
            --_level;
        }
        if (_rchild)
        {
            _rchild->print();
            --_level;
        }
    }
private:
    int _val;
    Node *_lchild;      
    Node *_rchild;
    static int _level;      
};

int Node::_level = 0;       

class Tree{
public:
    Tree(): _root(0){}  
    void add(int val)
    {
        Node *temp = new Node(val);
        if (!_root)
            _root = temp;
        else
            _root->add(temp);       
    }
    void print()
    {
        if (!_root)
            return;
        _root->print();             
    }
private:
    Node *_root;    
};

int main()
{
    Tree tree;
    tree.add(5);
    tree.add(9);
    tree.add(1);
    tree.add(7);
    tree.add(5);
    tree.add(14);
    tree.print();
}

我认为您不应该自己编写代码，而是看看Tree :: Visualize这似乎是一个不错的Perl实现，具有不同的可能样式，并在那里使用/移植一种算法。

我有一个Ruby程序，该程序计算应在此处绘制二叉树中的每个节点的坐标：http : //hectorcorrea.com/Blog/Drawing-a-Binary-Tree-in-Ruby

这段代码使用了非常基本的算法来计算坐标，虽然不是“高效区域”，但这是一个好的开始。如果您想查看代码“ live”，则可以在此处进行测试：http : //binarytree.heroku.com/