二叉树------层序遍历

来源：互联网发布：php网页编辑软件有哪些编辑：程序博客网时间：2024/05/01 19:10

题目描述：
给定一棵二叉树，要求从上到下按层输出。（每一层按从左到右的顺序）
思路：
按照按层创建二叉树的方法，使用一个队列。先将根节点入队，然后每次从队中取出队首元素，并将其左右子节点入队，持续此过程知道队列为空。（NULL不入队）

#include <iostream>#include <deque>#include <stdio.h>using namespace std;#define N 100struct Node {    char data;    struct Node *left, *right;};struct SeQueue{    struct Node *addr[N];    int front;    int end;};Node *MakeTree();void DisplayTree(Node *tree);void PrintTreeByFloor(Node *tree);void DeleteTree(Node **tree);int main(){    Node *myTree = MakeTree();    cout << "MidOrder: " << endl;    DisplayTree(myTree);    cout << endl;    cout << "print tree by floor: " << endl;    PrintTreeByFloor(myTree);    cout << endl;    DeleteTree(&myTree);    return 0;}Node *MakeTree(){    char ch;    SeQueue Q;    SeQueue *q = &Q;    q->front = 1;    q->end = 0;    Node *tmp = NULL;    Node *root = NULL;    cout << "Input some chars to make a tree(less than 99 and end by '@'):" << endl;    ch = getchar();    while(ch != '@' && ch != '\n')    {        tmp = new Node();        tmp->data = ch;        tmp->left = tem->right = NULL;        q->end += 1;        q->addr[q->end] = tmp;        if(root = NULL)        {            root = tmp;        }        else        {            if(q->addr[q->front] != NULL)            {                if(q->end % 2 == 0)                    q->addr[q->front]->left = tmp;                else                    q->addr[q->front]->right = tmp;            }            if(q->end % 2 != 0)                q->front += 1;        }        ch = getchar();    }    return root;}void DisplayTree(Node *tree){    Node *p = tree;    if(p)    {        DisplayTree(p->left);        cout << p->data << " ";        DisplayTree(p->right);    }}void PrintTreeByFloor(Node *tree){    if(tree == NULL)        return ;    deque<Node *> treeNode;    treeNode.push_back(tree);    while(treeNode.size())    {        Node *tmp = treeNode.front();        treeNode.pop_front();        cout << tmp->data << " ";        if(tmp->left)            treeNode.push_back(tmp->left);        if(tmp->right)            treeNode.push_back(tmp->right);    }}void DeleteTree(Node **tree){    if(tree == NULL)        return ;    Node *p = *tree;    if(p)    {        DeleteTree(&p->left);        DeleteTree(&p->right);        delete [] p;    }    *tree = NULL;}

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