怎样输出二叉树的所有路径（按字典序）？
@Author: 张海拔
@Update: 2014-01-30
@Link: http://www.cnblogs.com/zhanghaiba/p/3536534.html
二叉树的路径这里定义为：从树根到叶子的路径。
输出二叉树的所有路径，与“求n的全排列”如出一辙，都是使用回溯法的典范。
首先，先屏蔽掉“按字典序”输出的要求，看看怎样输出二叉树的所有路径。
print_all_path()函数，采用深度遍历（也是先序遍历）二叉树，当发现叶子节点意味着找到了一条路径，此时保存叶子的item到path，并输出path[]。
整个函数代码核心是cnt变量，代码要维护cnt，让它始终指向着path数组（可以看做栈）的下一个空闲位置（栈顶）。
所以哪里有回溯（退栈），哪里就需要执行cnt--;，以维护cnt变量。
现在，已经可以输出二叉树的所有路径了。
按字典序输出，在DFS选择扩展节点时，分三种情况处理就很清晰了：
若（1）左子树为空，则遍历右子树；
否则若（2）右子树为空，则遍历左子树；
否则（3），即左右子树均不为空，则判断左子树与右子树的根：若左根小则先遍历左子树后遍历右子树，否则先遍历右子树后遍历左子树。
以上面的思路改进print_all_path()，很容易写出print_all_path_dict_order()函数
下面的是完整代码实现及测试示范：
1 /* 2 *Author: ZhangHaiba 3 *Date: 2014-1-30 4 *File: print_all_path_of_binary_tree.c 5 * 6 *a demo shows how to print all path of binary tree in dictionary order 7 */ 8 9 #include <stdio.h> 10 #include <stdlib.h> 11 #include <limits.h> 12 #define LEN 1024 13 #define CMD_LEN 128 14 #define MOD 100 15 16 typedef struct node * link; 17 typedef struct node { 18 int item; 19 link left; 20 link right; 21 }node; 22 23 24 //public 25 link NODE(int item, link left, link right); 26 link bt_create_by_random(int n); 27 void bt_show_by_tree(link root); 28 void print_all_path(link root); 29 void print_all_path_dict_order(link root); 30 31 //private 32 void tree_print(link root, FILE *fd); 33 link bt_create_by_random_core(int n); 34 35 //private 36 int path[LEN]; 37 int cnt = 0; 38 39 int main(void) 40 { 41 int n; 42 43 scanf("%d", &n); 44 link bt_tree_a = bt_create_by_random(n); 45 bt_show_by_tree(bt_tree_a); 46 printf("print all path in DFS order:\n"); 47 print_all_path(bt_tree_a); 48 printf("print all path in dictionary order:\n"); 49 print_all_path_dict_order(bt_tree_a); 50 return 0; 51 } 52 53 54 void print_all_path(link root) 55 { 56 if (root == NULL) 57 return; 58 if (root->left == NULL && root->right == NULL) { 59 path[cnt++] = root->item; 60 int i; 61 for (i = 0; i < cnt; ++i) 62 printf(i == cnt-1 ? "%d\n" : "%d ", path[i]); 63 cnt--; 64 return; 65 } 66 path[cnt++] = root->item; 67 print_all_path(root->left); 68 print_all_path(root->right); 69 cnt--; 70 } 71 72 void print_all_path_dict_order(link root) 73 { 74 if (root == NULL) 75 return; 76 if (root->left == NULL && root->right == NULL) { 77 path[cnt++] = root->item; 78 int i; 79 for (i = 0; i < cnt; ++i) 80 printf(i == cnt-1 ? "%d\n" : "%d ", path[i]); 81 cnt--; 82 return; 83 } 84 path[cnt++] = root->item; 85 if (root->left == NULL) 86 print_all_path_dict_order(root->right); 87 else if (root->right == NULL) 88 print_all_path_dict_order(root->left); 89 else { 90 if (root->left->item < root->right->item) { 91 print_all_path_dict_order(root->left); 92 print_all_path_dict_order(root->right); 93 } else { 94 print_all_path_dict_order(root->right); 95 print_all_path_dict_order(root->left); 96 } 97 } 98 cnt--; 99 } 100 101 102 link NODE(int item, link left, link right) 103 { 104 link born = malloc( sizeof (node) ); 105 born->item = item; 106 born->left = left; 107 born->right = right; 108 return born; 109 } 110 111 link bt_create_by_random(int n) 112 { 113 srand( (unsigned)time(NULL) ); 114 return bt_create_by_random_core(n); 115 } 116 117 link bt_create_by_random_core(int n) 118 { 119 if (n <= 0) 120 return NULL; 121 if (n == 1) 122 return NODE(rand()%MOD, NULL, NULL); 123 link root = NODE(rand()%MOD, NULL, NULL); 124 int left_n = rand()%(n-1)+1, right_n = (n-1) - left_n; 125 root->left = bt_create_by_random_core(left_n); 126 root->right = bt_create_by_random_core(right_n); 127 return root; 128 } 129 130 131 void bt_show_by_tree(link root) 132 { 133 char cmd[CMD_LEN]; 134 135 sprintf(cmd, "rm -f ./tree_src.txt"); 136 system(cmd); 137 138 FILE *fd = fopen("./tree_src.txt", "a+"); 139 fprintf(fd, "\n\t\\tree"); 140 tree_print(root, fd); 141 fprintf(fd, "\n\n"); 142 fclose(fd); 143 144 sprintf(cmd, "cat ./tree_src.txt | ~/tree/tree"); 145 system(cmd); 146 } 147 148 void tree_print(link root, FILE *fd) 149 { 150 fprintf(fd, "("); 151 if (root != NULL) { 152 fprintf(fd, "%d", root->item); 153 tree_print(root->left, fd); 154 tree_print(root->right, fd); 155 } 156 fprintf(fd, ")"); 157 }
测试示范：
ZhangHaiba-MacBook-Pro:code apple$ ./a.out 15 57 ________|________ | | 76 98 _|__ _____|_____ | | | | 49 0 78 ___|___ ___|___ _|__ | | | | | | 24 10 96 22 97 _|__ _|__ _|__ _|__ _|__ | | | | | | | | | | 29 86 79 _|__ _|__ _|__ | | | | | | 73 _|__ | | print all path in DFS order: 57 76 49 24 57 76 49 10 29 57 98 0 96 86 73 57 98 0 22 79 57 98 78 97 print all path in dictionary order: 57 76 49 10 29 57 76 49 24 57 98 0 22 79 57 98 0 96 86 73 57 98 78 97 ZhangHaiba-MacBook-Pro:code apple$ ./a.out 15 85 ________|_________ | | 27 38 ___|___ ______|______ | | | | 75 81 64 39 _|__ _|__ _|__ ___|___ | | | | | | | | 28 27 50 28 _|__ ___|___ _|__ _|__ | | | | | | | | 44 30 73 _|__ _|__ _|__ | | | | | | 0 _|__ | | print all path in DFS order: 85 27 75 85 27 81 28 85 38 64 27 44 0 85 38 64 27 30 85 38 39 50 73 85 38 39 28 print all path in dictionary order: 85 27 75 85 27 81 28 85 38 39 28 85 38 39 50 73 85 38 64 27 30 85 38 64 27 44 0 ZhangHaiba-MacBook-Pro:code apple$ ./a.out 20 55 ________|_________ | | 81 39 ___|___ ______|______ | | | | 93 7 76 14 _|__ _|__ ___|___ ___|___ | | | | | | | | 79 24 98 41 20 _|__ _|__ _|__ _|__ _|__ | | | | | | | | | | 43 ____|____ | | 22 92 ___|____ _|__ | | | | 28 16 ___|___ _|__ | | | | 15 46 _|__ _|__ | | | | 71 _|__ | | print all path in DFS order: 55 81 93 55 81 7 79 55 39 76 24 55 39 76 98 55 39 14 41 55 39 14 20 43 22 28 15 71 55 39 14 20 43 22 28 46 55 39 14 20 43 22 16 55 39 14 20 43 92 print all path in dictionary order: 55 39 14 20 43 22 16 55 39 14 20 43 22 28 15 71 55 39 14 20 43 22 28 46 55 39 14 20 43 92 55 39 14 41 55 39 76 24 55 39 76 98 55 81 7 79 55 81 93
