PAT甲级专题|树的遍历

PAT甲级专题-树的遍历

涉及知识点：树、建树、深度优先搜索、广度优先搜索、递归

甲级PTA 1004

输出每一层的结点，邻接表vector建树后、用dfs、bfs都可以边搜边存当前层的数据，

#include
using namespace std;

const int maxn = 110;
int n, m;
vector g[maxn];
int ans[maxn];
int deep = 0;

void dfs(int x, int depth) {
    if (g[x].size() == 0) {
        if (depth > deep) deep = depth;
        ans[depth]++;
        return;
    }
    for (int i = 0; i < g[x].size(); i++) {
        dfs(g[x][i], depth + 1);
    }
}


int main() {
    cin >> n >> m;
    for (int i = 1; i <= m; i++) {
        int id,k;
        cin >> id;
        cin >> k;
        for (int j = 1; j <= k; j++) {
            int id2;
            cin >> id2;
            g[id].push_back(id2);
        }
    }
    dfs(1, 0);
    for (int i = 0; i <= deep; i++) {
        if (i != deep) cout << ans[i] << " ";
        else cout << ans[i];
    }
    return 0;
}

甲级PTA 1020

中序、后序序列，找出层次遍历的序列

前置知识，中序后序序列来建树、中序后序序列找出前序序列

//中序 后序  找 前序
/*
void build(int root, int start, int end) {
    if (start > end) return;
    if (root < 1) return;
    int pos = start;
    while (pos < end && iorder[pos] != porder[root]) pos++;
    pre[++idx] = porder[root];
    //cout << porder[root] << endl;
    build(root - (end - pos + 1), start, pos - 1);
    build(root - 1, pos + 1, end);
}
*/

本题代码

#include
using namespace std;

const int maxn = 50;
int n;
int porder[maxn];
int iorder[maxn];
int a[maxn];
int idx = 0;
int deep = 0;
vector v[maxn];
int ans[maxn];

//后序 中序 在递归时 使用vector存下当前一层的数，因为按左子树优先递归，所以存的序列符合题意
void build(int root, int start, int end, int depth) {
    if (start > end) return;
    int pos = start;
    while (pos < end && iorder[pos] != porder[root]) pos++;
    v[depth].push_back(porder[root]);
    if (depth > deep) deep = depth;
    //cout << "depth = " << depth << " " << porder[root] << endl;
    build(root - (end - pos + 1), start, pos - 1, depth + 1);
    build(root - 1, pos + 1, end, depth + 1);
}

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> porder[i];
    for (int i = 1; i <= n; i++) cin >> iorder[i];
    //build(n, 1, n, 1, n);
    //build(n, 1, n);
    build(n, 1, n, 1);
    int num = 1;
    for (int i = 1; i <= deep; i++) {
        for (int j = 0; j < v[i].size(); j++) {
            ans[num++] = v[i][j];
        }
    }
    for (int i = 1; i <= n; i++) {
        if (i == n) cout << ans[i];
        else cout << ans[i] << " ";
    }
    return 0;
}

甲级PTA 1053

找出路径长度等于 S 的路径，按结点权值的字典序排列
vector按权值大的优先存树的边、dfs搜索边搜边存答案。

#include
using namespace std;

typedef long long ll;
const int maxn = 1010;
vector g[maxn];
int w[maxn];
int n, m;
ll s;
int path[maxn];
vector ans[maxn];
int len = 0;

struct node {
    int v;
};

void dfs(int x, int depth) {
    path[depth] = w[x];
    if (g[x].size() == 0) {
        ll temp = 0;
        for (int i = 0; i <= depth; i++) temp += path[i];
        if (temp == s) {
            for (int i = 0; i <= depth; i++) ans[len].push_back(path[i]);
            len++;
        }
        return;
    }
    for (int i = 0; i < g[x].size(); i++) {
        dfs(g[x][i], depth + 1);
    }
}

bool cmp(int a,int b) {
    return w[a] > w[b];
}

node temp[1010];
int main() {
    cin >> n >> m >> s;
    for (int i = 0; i < n; i++) cin >> w[i];
    for (int i = 1; i <= m; i++) {
        int id1, k, id2;
        cin >> id1 >> k;
        for (int j = 1; j <= k; j++) {
            cin >> id2;
            g[id1].push_back(id2);
        }
    }
    for (int i = 0; i < n; i++) sort(g[i].begin(), g[i].end(), cmp);

    dfs(0, 0);

    for (int i = 0; i < len; i++) {
        for (int j = 0; j < ans[i].size(); j++) {
            if (j == ans[i].size() - 1) cout << ans[i][j] << endl;
            else cout << ans[i][j] << " ";
        }
    }

    return 0;
}

甲级PTA 1079

算出，从各个叶节点 到 根的距深度（dfs），按题意计算答案。

#include
using namespace std;

const int maxn = 1e5+10;
int n;
double r, p,ans = 0;
vector g[maxn];
double dat[maxn];

void dfs(int x, int depth) {
    if (g[x].size() == 0) {
        ans += (dat[x] * pow(1 + r, depth));
        return;
    }
    for (int i = 0; i < g[x].size(); i++) {
        dfs(g[x][i], depth + 1);
    }
}

int main() {
    cin >> n >> p >> r;
    r = r * 0.01;
    for (int i = 0; i < n; i++) {
        int k;
        cin >> k;
        int id;
        double val;
        if (k == 0) {
            cin >> val;
            dat[i] = val;
            continue;
        }
        for (int j = 0; j < k; j++) {
            cin >> id;
            g[i].push_back(id);
        }
    }
    dfs(0, 0);
    printf("%.1f", ans * p);
    return 0;
}

甲级PTA 1086

先序、中序序列，找出后序序列
我的做法：先根据先序中序序列建树、再后序遍历找出后序序列。
先序确定根，中序划分左右子树

但是这题题目没说完善，万一有重复元素，应该就建不了树了

#include
using namespace std;

const int maxn = 50;
int n;
struct node {
    int v;
    node * l, * r;
};

int pre[maxn];
int in[maxn];
int post[maxn];
int idx1 = 0, idx2 = 0,idx = 0;
stack st;

node * build(int root,int il,int ir) {
    if (il > ir) return NULL;
    int pos = il;
    while (pos <= ir && in[pos] != pre[root]) pos++;
    node* Root = new node;
    Root->v = pre[root];
    Root->l = build(root+1, il, pos - 1);
    Root->r = build(root+(pos-il)+1 ,pos+1,ir);
    return Root;
}

void postOrder(node *Root) {
    if (Root == NULL) return;
    if (Root->l) postOrder(Root->l);
    if (Root->r) postOrder(Root->r);
    post[++idx] = Root->v;
}

int main() {
    cin >> n;
    while (1) {
        string ins;
        int id;
        cin >> ins;
        if (ins == "Push") {
            cin >> id;
            pre[++idx1] = id;
            st.push(id);
        }else {
            int top = st.top();
            in[++idx2] = top;
            st.pop();
        }
        if (idx1 == n && idx2 == n) break;
    }
    node* Root = new node;
    Root = build(1, 1, n);
    postOrder(Root);
    for (int i = 1; i <= n; i++) {
        if (i == n) cout << post[i];
        else cout << post[i] << " ";
    }

    return 0;
}