【PAT】A1020. Tree Traversals (25)

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原创最新推荐文章于 2026-09-07 11:30:18 发布 275 阅读 AI 写同款· GEO 优化 ›

基于ZYNQ PL的AD7606数据采集与FFT实时频谱分析：从硬件到创新实现外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-kKnmpgJm-1750295287594)(https://example.com/fft_plot.png)]本文涉及代码已在Xilinx ZCU102开发板与AD7606评估板上验证通过，采样率支持200KSPS@8通道，FFT计算延迟<1ms。传统MCU+DSP方案难以兼顾速度与功耗，而ZYNQ的PL+PS架构为这一问题提供了完美解决方案。本文实现的ZYNQ PL方案突破了传统处理器的性能瓶颈，通过。阅读详情

A1020. Tree Traversals (25)

#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;

struct treeNode{
    int data;
    treeNode* lchild;
    treeNode* rchild;
};

treeNode* buildTree(int inl, int inr, int postl, int postr, int* inorder, int* postorder){
    if(postl > postr)
        return NULL;

    treeNode* root = new treeNode;
    root->data = postorder[postr];
    int rootIndex;
    for(int i = inl; i <= inr; i++){
        if(inorder[i] == root->data){
            rootIndex = i;
            break;
        }
    }
    root->lchild = buildTree(inl, rootIndex - 1, postl, postl + rootIndex - inl - 1, inorder, postorder);
    root->rchild = buildTree(rootIndex + 1, inr, postl + rootIndex - inl, postr - 1, inorder, postorder);
    return root;
}

void BFS(treeNode* root, int N){
    queue<treeNode*> q;
    q.push(root);
    int count = 0;
    while (q.size() != 0) {
        treeNode* node = q.front();
        q.pop();
        printf("%d", node->data);
        count++;
        if(count < N)
            printf(" ");
        if(node->lchild)
            q.push(node->lchild);
        if(node->rchild)
            q.push(node->rchild);
    }
}

int main(){
    int N, i;
    scanf("%d", &N);
    int inorder[N], postorder[N];
    for(i = 0; i < N; i++){
        scanf("%d", &postorder[i]);
    }
    for(i = 0; i < N; i++){
        scanf("%d", &inorder[i]);
    }

    treeNode* root = buildTree(0, N - 1, 0, N - 1, inorder, postorder);
    BFS(root, N);

    return 0;
}