題目鏈接:https://pintia.cn/problem-sets/994805342720868352/problems/994805407749357568
題目大意:給出n個數,讓你建立一棵完全二叉樹,並輸出層序遍歷的結果。
分析:先把給出的數組排序,再求出要建立的樹有多少層。因爲中序遍歷的結果就是順序的,所以按照中序遞歸,每遞歸一層,層數減一,求出當前根節點是第幾個數,然後先往左子樹走,再給當前根節點賦值,最後往右子樹走。當只剩下一個節點或者層數爲1時,賦值,返回。最後層序遍歷一下就可以了。
數組開大點。。。
#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 5;
int n, pos, a[N], ans[N], sum[N];
void init() {
sum[1] = 1;
for(int i = 2; i < 15; i++)
sum[i] = sum[i - 1] + pow(2, i - 1);
}
void dfs(int x, int cnt, int l, int r) {
if(l > r) return ;
if(l == r || cnt == 1) {
ans[x] = a[pos++];
return ;
}
int num = r - l + 1;
int rt = (sum[cnt - 1] - 1) / 2;
int cha = sum[cnt] - sum[cnt - 1];
rt += min(num - sum[cnt - 1], cha / 2);
rt++;
rt = rt + l - 1;
dfs(x<<1, cnt - 1, l, rt - 1);
ans[x] = a[rt];
pos++;
dfs(x<<1|1, cnt - 1, rt + 1, r);
}
void solve() {
queue<int> q;
q.push(1);
int d = 0;
while(!q.empty()) {
int tmp = q.front();
q.pop();
if(d > 0) printf(" ");
printf("%d", ans[tmp]);
d++;
if(ans[tmp<<1] != -1) q.push(tmp<<1);
if(ans[tmp<<1|1] != -1) q.push(tmp<<1|1);
}
printf("\n");
}
int main() {
init();
scanf("%d", &n);
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
sort(a + 1, a + n + 1);
int cnt = 1;
while(n >= sum[cnt]) cnt++;
if(n - sum[cnt] == 0) cnt--;
pos = 1;
memset(ans, -1, sizeof ans);
dfs(1, cnt, 1, n);
solve();
return 0;
}
鏈表實現
#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int n, pos, a[N], ans[N], sum[N];
struct node {
int data;
node *left, *right;
};
struct node *root = NULL;
void init() {
sum[1] = 1;
for(int i = 2; i < 15; i++)
sum[i] = sum[i - 1] + pow(2, i - 1);
}
void dfs(node * &root, int cnt, int l, int r) {
if(l > r) return ;
if(l == r || cnt == 1) {
root = new node();
root->data = a[pos++];
root->left = root->right = NULL;
return ;
}
if(root == NULL) {
root = new node();
root->left = root->right = NULL;
}
int num = r - l + 1;
int rt = (sum[cnt - 1] - 1) / 2;
int cha = sum[cnt] - sum[cnt - 1];
rt += min(num - sum[cnt - 1], cha / 2);
rt++;
rt = rt + l - 1;
dfs(root->left, cnt - 1, l, rt - 1);
root->data = a[rt];
pos++;
dfs(root->right, cnt - 1, rt + 1, r);
}
void solve() {
queue<node*> q;
q.push(root);
int d = 0;
while(!q.empty()) {
node *tmp = q.front();
q.pop();
if(d > 0) printf(" ");
printf("%d", tmp->data);
d++;
if(tmp->left != NULL) q.push(tmp->left);
if(tmp->right != NULL) q.push(tmp->right);
}
printf("\n");
}
int main() {
init();
scanf("%d", &n);
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
sort(a + 1, a + n + 1);
int cnt = 1;
while(n >= sum[cnt]) cnt++;
if(n - sum[cnt] == 0) cnt--;
pos = 1;
dfs(root, cnt, 1, n);
solve();
return 0;
}