UVa122 二叉树动态创建与BFS

这题就是定义了二叉树的结点类型，然后由根节点开始通过解析表达式LR……生成一个树
注意判断一个结点是否被多次赋值，题目要求一个节点多次赋值最后答案要输出错误结论，提交的时候曾经因为这个问题WA了
树的BFS：通过队列实现，一开始先把根节点push，当队列不为空时取出队列头保存，把它的左子结点和右子节点push进队，就可以完成树的层次遍历

代码实现

#include <iostream>
#include <string>
#include <vector>
#include <queue>
using namespace std;

//树的节点
struct node
{
    bool have_value;
    int value;
    node *left;
    node *right;
    node() {
        have_value = false;
        left = NULL;
        right = NULL;
    }
};
node *root; //根节点
bool failed;
//input->addnode->bfs
void addnode(int value, string LR)
{
    int n = LR.size();
    if (n == 2)
    {
        root->value = value;
        root->have_value = true;
    }
    else
    {
        node *temp=root;
        for (int i = 1; i < n; i++)
        {
            if (LR[i] == 'L')
            {
                if (temp->left == NULL)
                    temp->left = new node();
                temp = temp->left;
            }
            else if (LR[i] == 'R')
            {
                if (temp->right == NULL)
                    temp->right = new node();
                temp = temp->right;
            }
        }
        //找到要赋值的节点，准备赋值
        if (temp->have_value)
            failed = true;
        temp->value = value;
        temp->have_value = true;
        //cout << temp->value << endl;
    }
}
bool input()
{
    string inputOrder;
    failed = false;//判断是否有重复的值
    root = new node();
    for(;;)
    {
        if (!(cin >> inputOrder))
            return false;
        if (inputOrder.size() == 2) //()的情况
            break;
        int value= stoi(inputOrder.substr(inputOrder.find_first_of("0123456789")));
        //cout << value;
        string LR = inputOrder.substr(inputOrder.find(","));
        //cout << " " << LR << endl;
        addnode(value, LR);
    }
    return true;
}
bool bfs(vector<int> &result) //如果有节点没有被赋值则返回false
{
    result.clear();
    node *temp = root;
    queue<node*> bfsqueue;
    if (root->have_value)
        bfsqueue.push(temp);
    else
        return false;
    while (!bfsqueue.empty())
    {
        node *front = bfsqueue.front();
        if (front->have_value)
            result.push_back(front->value);
        else
            return false;
        bfsqueue.pop();
        if (front->left != NULL)
        {
            bfsqueue.push(front->left);
        }
        if (front->right != NULL)
        {
            bfsqueue.push(front->right);
        }
    }
    return true;
}
int main()
{
    while (input())
    {
        vector<int> result;
        if (bfs(result)&&(!failed))
        {
            for (auto it = result.begin(); it != result.end(); it++)
            {
                if (it == result.begin())
                    cout << *it;
                else
                    cout << " "<<*it;
            }
            cout << endl;
        }
        else
        {
            cout << "not complete" << endl;
        }
    }
    system("pause");
    return 0;
}