有哪些算法可以在控制台中绘制二叉树?该树是用C实现的。例如,具有数字2 3 4 5 8的BST将在控制台中显示为:
有哪些算法可以在控制台中绘制二叉树?该树是用C实现的。例如,具有数字2 3 4 5 8的BST将在控制台中显示为:
代码:
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)
或者 (006)
+------------------------+---------+
(001) (008)
+----+---------+ +----+----+
(-02) (003) (007) (009)
+----+ +----+---------+
(-06) (002) (005)
+---------+ +----+
(-08) (004)
+----+----+
(-09) (-07)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!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
这是一种将树实现在数组中的方法:
#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
#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;
}
0
1 2
3 4 5 6
node_depth
。也许你复制时打错了?这里是可工作的代码:http://ideone.com/wrY8Vo - Steve Lorimer我赞同litb的建议。最近我也不得不这样做,以打印Windows进程的VAD树,并使用DOT语言(只需从二叉树遍历函数中打印节点):
http://en.wikipedia.org/wiki/DOT_language
例如,您的DOT文件将包含:5
1
5
9
7
14
代码(Node::print()
函数是最重要的):
#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();
}