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<bits/stdc++.h>
using namespace std;

const int maxn = 110;
int n, m;
vector<int> 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<bits/stdc++.h>
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<int> 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<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn = 1010;
vector<int> g[maxn];
int w[maxn];
int n, m;
ll s;
int path[maxn];
vector<int> 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<bits/stdc++.h>
using namespace std;

const int maxn = 1e5+10;
int n;
double r, p,ans = 0;
vector<int> 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<bits/stdc++.h>
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<int> 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;
}