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/*
Challenge 094
TASK #2 � Binary Tree to Linked List
Submitted by: Mohammad S Anwar
You are given a binary tree.
Write a script to represent the given binary tree as an object and flatten
it to a linked list object. Finally print the linked list object.
Example:
Input:
1
/ \
2 3
/ \
4 5
/ \
6 7
Output:
1 -> 2 -> 4 -> 5 -> 6 -> 7 -> 3
*/
#include <assert.h>
#include <ctype.h>
#include <memory.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAXLINE 256
// check memory allocation
void* check_mem(void* p) {
if (!p) {
fputs("out of memory", stderr);
exit(EXIT_FAILURE);
}
return p;
}
// tree object
typedef struct Tree {
int value;
struct Tree* left, * right;
} Tree;
// input image
char** lines;
size_t num_lines;
// create and delete a tree
Tree* tree_new() {
Tree* node = check_mem(calloc(1, sizeof(Tree)));
return node;
}
void tree_delete(Tree* node) {
if (node->left) tree_delete(node->left);
if (node->right) tree_delete(node->right);
free(node);
}
// read lines from stdin
void lines_read() {
char line[MAXLINE];
while (fgets(line, sizeof(line), stdin)) {
lines = check_mem(realloc(lines, (num_lines + 1) * sizeof(char*)));
lines[num_lines] = check_mem(strdup(line));
num_lines++;
}
}
void lines_delete() {
if (lines) {
for (size_t i = 0; i < num_lines; i++)
free(lines[i]);
free(lines);
}
}
// parse the tree
Tree* parse_subtree(int row, int col) {
// parse root
Tree* node = tree_new();
assert(isdigit(lines[row][col]));
node->value = lines[row][col] - '0';
// parse children
if (row + 2 < (int)num_lines) {
// parse left subtree
if (col - 2 >= 0 && lines[row + 1][col - 1] == '/')
node->left = parse_subtree(row + 2, col - 2);
// parse right subtree
if (col + 2 < (int)strlen(lines[row + 2]) && lines[row + 1][col + 1] == '\\')
node->right = parse_subtree(row + 2, col + 2);
}
return node;
}
Tree* parse_tree() {
assert(num_lines > 0);
char* p = strpbrk(lines[0], "0123456789");
assert(p);
int col = p - lines[0];
return parse_subtree(0, col);
}
/*
// compute tree depth
void subtree_depth(Tree* node, int cur_depth, int* pmax_depth) {
if (!node)
return;
if (node->left)
subtree_depth(node->left, cur_depth + 1, pmax_depth);
if (node->right)
subtree_depth(node->right, cur_depth + 1, pmax_depth);
if (cur_depth > *pmax_depth)
*pmax_depth = cur_depth;
}
int tree_depth(Tree* root) {
int max_depth = 0;
subtree_depth(root, 1, &max_depth);
return max_depth;
}
// print a tree
void print_subtree(Tree* node, char* lines[], int row, int col) {
if (!node)
return;
lines[row][col] = '0' + node->value;
if (node->left) {
lines[row + 1][col - 1] = '/';
print_subtree(node->left, lines, row + 2, col - 2);
}
if (node->right) {
lines[row + 1][col + 1] = '\\';
print_subtree(node->right, lines, row + 2, col + 2);
}
}
void print_tree(Tree* root) {
// create canvas
int depth = tree_depth(root);
int num_lines = depth * 2;
int num_cols = depth * 5;
char** lines = check_mem(calloc(num_lines, sizeof(char*)));
for (int i = 0; i < num_lines; i++) {
lines[i] = check_mem(calloc(num_cols + 1, 1));
memset(lines[i], ' ', num_cols);
lines[i][num_cols] = '\0';
}
// draw tree
print_subtree(root, lines, 0, num_cols / 2);
// print tree
for (int i = 0; i < num_lines; i++)
printf("%s\n", lines[i]);
// delete canvas
for (int i = 0; i < num_lines; i++)
free(lines[i]);
free(lines);
}
*/
// pre-order traversal, set left to NULL, move subtree to right
Tree* flatten(Tree* root) {
if (!root) return root;
Tree* left = flatten(root->left);
Tree* right = flatten(root->right);
root->left = NULL;
root->right = left;
Tree* node = root;
while (node->right) // find right-most node
node = node->right;
node->right = right;
return root;
}
int main() {
lines_read();
Tree* tree = parse_tree();
flatten(tree);
for (Tree* node = tree; node; node = node->right)
printf("%d%s", node->value, node->right ? " -> " : "\n");
tree_delete(tree);
lines_delete();
}
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