1138. Postorder Traversal (25)

1138. Postorder Traversal (25)

Suppose that all the keys in a binary tree are distinct positive integers. Given the preorder and inorder traversal sequences, you are supposed to output the first number of the postorder traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=50000), the total number of nodes in the binary tree. The second line gives the preorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the first number of the postorder traversal sequence of the corresponding binary tree.

Sample Input:

71 2 3 4 5 6 72 3 1 5 4 7 6

Sample Output:

3

解析：

根据前序和中序排列，生成一棵树，后序遍历生成的二叉树，将结果保存到post数组中，输出post数组第一个数。

代码如下：

#include<iostream>#include<vector>using namespace std;vector<int> pre, in,post;struct node {    int data;    struct node *left,*right;};node *CreateTree(node *root,int i,int j,int k) {    if (i > j)return NULL;    root = new node;    root->left = root->right = NULL;    root->data = pre[i];    int temp = pre[i];    int t = k;    while (in[t] != temp)t++;    int len = t - k;    root->left = CreateTree(root->left, i + 1, i + len, k);    root->right = CreateTree(root->right, i + len+1, j, t + 1);    return root;}int i = 0;void last(node *root) {    if (root == NULL)return;    last(root->left);    last(root->right);    post[i++]=root->data;}int main(){    int N;    cin >> N;    pre.resize(N);    in.resize(N);    post.resize(N);    for (int i = 0; i < N; i++)cin >> pre[i];    for (int i = 0; i < N; i++)cin >> in[i];    node *root = NULL;    root = CreateTree(root, 0, N - 1, 0);    last(root);    cout << post[0];    return 0;}