树的层次遍历

来源:互联网 发布:思维脑图软件 编辑:程序博客网 时间:2024/05/01 07:56
#include<iostream>#include<queue>using namespace std;//结点权值作为结点编号int postOrder[31];     //后序遍历结点int inOrder[31];       //中序遍历结点int leftNodes[31];              //保存某结点的左子树编号int rightNodes[31];           //保存某结点的右子树编号//根据inOrder[L1]到inOrder[R1]  和postOrder[L1]到postOrder[R1]的结点编号 来构建树//返回根节点int buildTree(int L1, int R1, int L2, int R2){    if (R1 < L1)  //空树        return -1;    int root = postOrder[R2];      //后序遍历序列最后一个结点一定是根结点    int p =0;    while (inOrder[p] != root)    //找到中序遍历序列中对应哪个根结点的结点        p++;    int count = p - L1;          //左子树结点总数    //p是中序序列的根,从L1到p-1为左子树,对应的后续序列的从L2到L2+count-1    leftNodes[root] = buildTree(L1, p- 1, L2, L2 + count - 1);       //中序序列从p+1到R1为右子树,对应的后续序列从L2+count到R2 - 1 !!因为根节点已经去掉了!!    rightNodes[root] = buildTree(p + 1, R1, L2+count, R2-1);    return root;}//层序遍历//传了个N进去是因为输出格式控制 = = void printVex(int root,int N){     queue<int> q;    q.push(root);    while (q.size()){        int vex = q.front();        if (N==1)            cout << vex;        else            cout << vex<<" ";        q.pop();        if (leftNodes[vex] != -1)            q.push(leftNodes[vex]);        if (rightNodes[vex] != -1)            q.push(rightNodes[vex]);        N--;    }}int main(){    int N;    cin >> N;    int index = 0;    for (int i = 1; i <= N; i++){        int vex;        cin >> vex;        postOrder[index++] = vex;    }    index = 0;    for (int j = 1; j <= N; j++){        int vex;        cin >> vex;        inOrder[index++] = vex;    }    int root = buildTree(0, N - 1, 0, N - 1);    printVex(root,N);    return 0;}

0 0
原创粉丝点击