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单链表#include #include struct Node{ char ch; Node * next;};struct CharList{ Node * head; Node * tail; Node * prev;};void InitCharList(CharList * list){ list->head = list->tail = list->p

树的遍历，获取深度#include #define ADT floattypedef struct TNode{ struct TNode * parent;//父结点 struct TNode * lchild;//左子树 struct TNode * rchild;//右子树 ADT value;} * BinTree;void SetTNode(TNode * p

#include #include #define ROW 6#define COL 6int a[ROW][COL] = { 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1};//1：通路； 0：障碍s

先序遍历：A->B->D->E->C中序遍历：D->B->E->A->C后序遍历：D->E->B->C->A#include typedef char ElemType;#define EndElem '^'//用^来表示空typedef struct BiNode{ ElemType elem; struct BiNode * lchild; struct B

1. 静态的示意  由N个顶点的无向连通图生成最小生成树的各条边放在容量为N - 1的数组中。#include #include #define N 6//N个顶点struct Edge {int i; int j; int w;};//i: 起点下标 j: 终点下标 w: 权//int (*adjmatrix)[N]: 邻接矩阵行地址//Edge edge[N - 1]: 输

#include #define N 6// 深度优先遍历void DFS_Traverse(bool adjmatrix[][N], int v0, void (*f)(int)){ bool visited[N] = {true};// v0设为已访问 int stack[N] = {v0};// 栈添加v0 int size = 1; f(v0); while (size