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:7 1 2 3 4 5 6 7 2 3 1 5 4 7 6Sample Output:
3
利用前序和中序序列建立二叉树，然后后序遍历二叉树，只输出一个结点即可。
利用前序和中序序列建立二叉树采用递归思路。首先根据前序序列的第一个元素作为根节点的值，再根据第一个元素将中序序列分为左右两段，然后对左字段递归建立二叉树作为根节点的左子树，对右字段递归建立二叉树作为根节点的右子树即可。
#include <cstdio>
#define MAX 50000struct Node{int data;Node *left, *right;
};
int pre[MAX], in[MAX];
int n;
bool judge = true;Node* create(int prel, int prer, int inl, int inr){if(prel > prer) return NULL;Node* root = new Node;root->data = pre[prel];int k = -1;for(int i = inl; i <= inr; i++){if(in[i] == root->data){k = i;break;}}if(k == -1) return NULL;int lnum = k - inl;root->left = create(prel+1, prel+lnum, inl, k-1);root->right = create(prel+lnum+1, prer, k+1, inr);return root;
}void postorder(Node* root){if(root != NULL && judge){postorder(root->left);postorder(root->right);if(judge){printf("%d", root->data);judge = false;}}
}int main(){scanf("%d", &n);for(int i = 0; i < n; i++) scanf("%d", &pre[i]);for(int i = 0; i < n; i++) scanf("%d", &in[i]);Node* root = create(0, n-1, 0, n-1);postorder(root);return 0;
}