C/C++中怎么實(shí)現(xiàn)樹(shù)操作

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預(yù)處理命令

#include <stdio.h>#include <stdlib.h>#define TRUE 1#define FALSE 0typedef int elemtype;typedef struct tNode* tree;typedef struct tNode { elemtype elem; tree left; tree right;}tNode;

計(jì)算樹(shù)的節(jié)點(diǎn)個(gè)數(shù)

//明確函數(shù)的功能：返回傳入樹(shù)的節(jié)點(diǎn)個(gè)數(shù)//定好尾頭：尾：當(dāng)傳入的節(jié)點(diǎn)尾NULL時(shí) 頭：1 + count(t->left) + count(t->right)int count(tree t){ if (t == NULL) return 0; return 1 + count(t->left) + count(t->right);}

求樹(shù)中節(jié)點(diǎn)數(shù)據(jù)為num的節(jié)點(diǎn)個(gè)數(shù)

//明確函數(shù)功能：返回節(jié)點(diǎn)數(shù)據(jù)為num的節(jié)點(diǎn)個(gè)數(shù)//定好尾頭：尾：NULL 頭：1 + func(左) + func（右） // 或者 func(左) + func（右）int count_num(tree t, elemtype num){ if (t == NULL) return 0; else { if (t->elem == num) return 1 + count_num(t->left, num) + count_num(t->right, num); else return count_num(t->left, num) + count_num(t->right, num); }}

求樹(shù)中節(jié)點(diǎn)數(shù)據(jù)的總和

//明確函數(shù)功能：返回總和//定好尾頭：尾：NULL 頭：root-> elem + func(左) + func（右）int add(tree t){ if (t == NULL) return 0; else return t->elem + add(t->left) + add(t->right);}

判斷樹(shù)中有無(wú)數(shù)據(jù)為num的節(jié)點(diǎn)

//兩種方式：一種是可以達(dá)成目的就結(jié)束，一種是需要遍歷完全才結(jié)束//明確函數(shù)功能：判斷其中有沒(méi)有值為num的節(jié)點(diǎn)返回1或0//定好尾頭：尾：值為num ，頭：int inTree_1(tree t, elemtype num){ if (t->elem == num) return TRUE; else { if (t->left != NULL) intree(t->left, num); // 使用遞歸將其遞到子節(jié)點(diǎn) if (t->right != NULL) intree(t->right, num); } return FALSE;}//確定函數(shù)功能：根據(jù)num的有無(wú)，返回0/非0//定好尾頭：尾：NULL 頭：有:return 1 + func(左)+func(右) 無(wú)：func(左)+func(右)int inTree_2(tree t, elemtype num){ if (t == NULL) return 0; int res; if (t->elem == num) res = 1+ intree(t->left, num) + intree(t->right, num);  if (t->elem == num) res = intree(t->left, num) + intree(t->right, num); return res;}

計(jì)算值為num的個(gè)數(shù)

int count_elem(tree t, elemtype val, int* num){ int val_l, val_r; if (t->left == NULL) return t->elem; if (t->right == NULL) return t->elem; else { val_l = count_elem(t->left, val, num); if (val == val_l) (*num)++; val_r = count_elem(t->right, val, num); if (val == val_r) (*num)++; return t->elem; } return *num;}

打印trunk

//明確函數(shù)功能：打印trunk//定好尾頭 尾：NULL 頭：第一步是判斷本節(jié)點(diǎn)是否是樹(shù)干然后打印，再func（左）去打印左邊的樹(shù)干 func（右）去打印右邊的樹(shù)干void print_trunk(tree t){ if (t == NULL) return; if (t->right != NULL || t->left != NULL) printf("%d", t->elem); print_trunk(t->right); print_trunk(t->left);}

判斷兩棵樹(shù)是否一樣

int same(tree t1, tree t2){ if (count(t1) == count(t2)) { if (t1->elem != t2->elem) return FALSE; if (t1->left != NULL && t2->left != NULL) same(t1->left, t2->left); if (t1->right != NULL && t2->right != NULL) same(t1->right, t2->right); return TRUE; } else return FALSE;}

求樹(shù)的高度

#define max(x, y) (x > y) ? x : yint height(tree t){ if (t == NULL)return -1; return 1 + max(height(t->right), height(t->left));}

打印樹(shù)中某值的層數(shù)

//明確函數(shù)功能：尋找放入的數(shù)的層數(shù)并打印//確定尾：//找到特定值的節(jié)點(diǎn) 找到NULL 頭：若是則打印，若不是則去左右子樹(shù)尋找layer++，當(dāng)孩子尋找完都沒(méi)有時(shí)layer--bool flag = false; //flag標(biāo)記可以用于提前結(jié)束遞歸void getTreeLayer(Node * root, int num, int &layer){ if (root == NULL) return; if (flag == true) return; if (root->data == num) {  cout << "num值" << num << "的層數(shù)為：" << layer << endl;  flag = true;  return;  } layer++; getTreeLayer(root->lChild, num); getTreeLayer(root->rChild, num); layer--;}

求節(jié)點(diǎn)的路徑

vector<int> path;bool flag = false; //flag標(biāo)記可以用于提前結(jié)束遞歸void getTreeLayer(Node * root, int num, int &layer){ if (root == NULL) return; if (flag == true) return; if (root->data == num) {  for(int x : path) cout << x << " "; bool flag = true; return; } path.push_back(); getTreeLayer(root->lChild, num); getTreeLayer(root->rChild, num); path.pop_back();}

上述就是小編為大家分享的C/C++中怎么實(shí)現(xiàn)樹(shù)操作了，如果剛好有類似的疑惑，不妨參照上述分析進(jìn)行理解。如果想知道更多相關(guān)知識(shí)，歡迎關(guān)注億速云行業(yè)資訊頻道。