PAT 甲级 刷题日记|A 1127 ZigZagging on a Tree (30 分)

题目

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in "zigzagging order" -- that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.

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Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤30), the total number of nodes in the binary tree. The second line gives the inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the zigzagging sequence of the tree in a line. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1结尾无空行

Sample Output:

1 11 5 8 17 12 20 15结尾无空行

思路

中序加后序建树，z字形输出（bfs存储到分层，按奇偶层正逆序输出） 比较常见了

代码

#include 
using namespace std;

const int maxn = 32;
int n;
int inorder[maxn];
int postorder[maxn];
vector za[maxn];
int maxhei = 0;


struct node {
    int data;
    node* lchild;
    node* rchild;
    node(int d): data(d), lchild(NULL), rchild(NULL) {
    }
};

void dfs(node* root, int height) {
    if (root == NULL) return ;
    if (height > maxhei) maxhei = height;
    za[height].push_back(root->data);
    dfs(root->lchild, height + 1);
    dfs(root->rchild, height + 1);
}

node* create(int ina, int inb, int posta, int postb) {
    if (ina > inb) return NULL;
    int ro = postorder[postb];
    node* root = new node(ro);
    int i;
    for (i = ina; i <= inb; i++) {
        if (inorder[i] == ro) break;
    }
    int numleft = i - ina;
    root->lchild = create(ina, ina + numleft - 1, posta, posta + numleft - 1);
    root->rchild = create(ina + numleft + 1, inb, posta + numleft, postb - 1);
    return root;
}

int main() {
    cin>>n;
    for (int i = 0; i < n; i++) {
        cin>>inorder[i];
    }   
    for (int i = 0; i < n; i++) {
        cin>>postorder[i];
    }
    node* root = NULL;
    root = create(0, n - 1, 0, n - 1);
    dfs(root, 0); 
    cout<data;
    for (int i = 1; i <= maxhei; i++) {
        if (i % 2 == 1) {
            for (int j = 0; j < za[i].size(); j++) {
                cout<<" "<= 0 ; j--) {
                cout<<" "<