二叉树基本操作-C语言实现

#include <iostream>
#include <stdlib.h>
using namespace std;
typedef struct node
{
    char sh;
    struct node *lchild, *rchild;
}TreeNode, *pTreeNode;

//这里注意，必须要使用&，不然后面求二叉树深度时有误
void Creat_Tree(pTreeNode & T)
{
    char ch;
    cin >> ch;
    if (ch == '#')
        T = NULL;
    else
    {
        T = new TreeNode;
        T->sh = ch;
        Creat_Tree(T->lchild);
        Creat_Tree(T->rchild);
    }
}
void Before_Search(pTreeNode T)
{
    //二叉树先序遍历
    if (T)
    {
        cout << T->sh;
        Before_Search(T->lchild);
        Before_Search(T->rchild);
    }
}
void Medium_Search(pTreeNode T)
{
    //二叉树中序遍历
    if (T)
    {
        Medium_Search(T->lchild);
        cout << T->sh;
        Medium_Search(T->rchild);
    }
}
void Last_Search(pTreeNode T)
{
    //二叉树后序遍历
    if (T)
    {
        Last_Search(T->lchild);
        Last_Search(T->rchild);
        cout << T->sh;
    }
}
int Depth_Tree(pTreeNode T)
{
    int m, n;
    if (T == NULL)
        return 0;
    {
        m = Depth_Tree(T->lchild);
        n = Depth_Tree(T->rchild);
        if (m > n)
            return m + 1;
        else
            return n + 1;
    }
}
int NumNode_Tree(pTreeNode T)
{
    if (T == NULL)
        return 0;
    else
        return NumNode_Tree(T->lchild) + NumNode_Tree(T->rchild) + 1;
}
int Leaves_Num_Tree(pTreeNode T)
{
    //定义静态变量使之在函数返回前一直存在
    static int LeavesNum = 0;
    if (T == NULL)
        return 0;
    //如果左和右孩子同时为空，则返回叶子树目
    //否则继续访问该结点的左孩子和右孩子

    if (T->lchild == NULL && T->rchild == NULL)
        LeavesNum++;
    else
    {
        Leaves_Num_Tree(T->lchild);
        Leaves_Num_Tree(T->rchild);
    }
    return LeavesNum;
}
int Branch_1_Tree(pTreeNode T)
{
    static int Branch_1_Num = 0;
    if (T == NULL)   //都空
        return 0;
    //如果左孩子为空或者右孩子为空，则返回度为1的结点个数
    //否则继续访问该结点的左孩子和右孩子
    if ((T->lchild == NULL && T->rchild != NULL) || (T->lchild != NULL && T->rchild == NULL))   //一个为空
    {
        Branch_1_Num++;
    }
    if (T->lchild != NULL || T->rchild != NULL) 
    {
        //10、11、01
        //左孩子不空右孩子空、左右都空、左孩子空右孩子不空
        //因此递归调用的条件是左孩子和右孩子不都为空
        Branch_1_Tree(T->lchild);
        Branch_1_Tree(T->rchild);
    }
    return Branch_1_Num;
}
int Branch_2_Tree(pTreeNode T)
{
    static int Branch_2_Num = 0;
    if (T == NULL)   //都空
        return 0;
    if (T->lchild != NULL && T->rchild != NULL)   //一个为空
        Branch_2_Num++;
    //有一个为空时候进行递归调用
    if (T->lchild != NULL || T->rchild != NULL)
    {
        Branch_2_Tree(T->lchild);
        Branch_2_Tree(T->rchild);
    }
    return Branch_2_Num;
}
void AllPath(pTreeNode T)
{
    int i;
    //定义静态变量，使得其一直保存到函数调用结束为止
    static int pathlen = 0;
    static char path[100];
    if (T != NULL)
    {
        if (T->lchild == NULL&&T->rchild == NULL)
        {
            printf("%c到根节点的逆路径：%c ", T->sh, T->sh);
            for (i = pathlen - 1; i >= 0; i--)
                printf("%c ", path[i]);
            printf("\n");
        }
        else
        {
            path[pathlen] = T->sh;//将当前节点放入路径中
            pathlen++;//路径长度增1
            AllPath(T->lchild);
            AllPath(T->rchild);
            pathlen--;
        }
    }
}
void main()
{
    pTreeNode T = NULL;
    cout << "创建二叉树，输入字符：" ;
    Creat_Tree(T);
    int depth = Depth_Tree(T);
    cout << "二叉树的深度是：";
    cout << depth << endl;

    int num = NumNode_Tree(T);
    cout << "二叉树结点个数是：";
    cout << num << endl;

    int Leaves_Num = Leaves_Num_Tree(T);
    cout << "二叉树叶子个数是：";
    cout << Leaves_Num << endl;
    
    int Branch_1_Num = Branch_1_Tree(T);
    cout << "二叉树度为1的结点个数：";
    cout << Branch_1_Num << endl;

    int Branch_2_Num = Branch_2_Tree(T);
    cout << "二叉树度为2的结点个数：";
    cout << Branch_2_Num << endl;

    cout << endl;
    AllPath(T);
    system("pause");
}

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