PAT A1119 . Pre- and Post-order Traversals (30)

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, or preorder and inorder traversal sequences. However, if only the postorder and preorder traversal sequences are given, the corresponding tree may no longer be unique. Now given a pair of postorder and preorder traversal sequences, you are supposed to output the corresponding inorder traversal sequence of the tree. If the tree is not unique, simply output any one of them. 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 preorder 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, first printf in a line “Yes” if the tree is unique, or “No” if not. Then print in the next line the inorder traversal sequence of the corresponding binary tree. If the solution is not unique, any answer would do. It is guaranteed that at least one solution exists. 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 1: 7 1 2 3 4 6 7 5 2 6 7 4 5 3 1 Sample Output 1: Yes 2 1 6 4 7 3 5 Sample Input 2: 4 1 2 3 4 2 4 3 1 Sample Output 2: No 2 1 3 4 #include <stdio.h>#include #include using namespace std; vector ans; int pre, post; bool uni = true;//true中序树唯一 struct node{ int data; node left, right; }; node create(int preL, int preR, int postL, int postR){ if (preL > preR) { return NULL; } node root = new node; root->data = pre[preL]; root->left = root->right = NULL; if (preL == preR) { return root; } int k = 0; for (k = preL + 1; k <= preR; k++) { if (pre[k] == post[postR - 1]) { break;//定位根节点的孩子结点在先序中的位置 } } if (k - preL > 1) {//不是先序中的左孩子,唯一,确定是根的右孩子 int numLeft = (k - 1) - (preL + 1);//根的右孩子前面都是左子树 root->left = create(preL + 1, k - 1, postL, postL + numLeft); root->right = create(k, preR, postL + numLeft + 1, postR - 1); }else{//无法确定是左孩子还是右孩子 uni = false;//生成树不唯一 root->right = create(k, preR, postL, postR - 1);//作为右孩子 } return root; } void inOrder(node root,vector& vi){ if (root == NULL) { return; } inOrder(root->left, vi); vi.push_back(root->data); inOrder(root->right, vi); } int main(){ int n = 0; vector in; scanf(“%d”, &n); pre = new int[n]; post = new int[n]; for (int i = 0; i < n; i++) { scanf(“%d”, &pre[i]); } for (int i = 0; i < n; i++) { scanf(“%d”, &post[i]); } node root = create(0, n - 1, 0, n - 1); inOrder(root, in); if (uni) { printf(“Yes\n”); }else{ printf(“No\n”); } for (int i = 0; i < in.size(); i++) { printf(“%d”, in[i]); if (i < in.size() - 1) { printf(“ “); }else{ printf(“\n”); } } return 0; }</stdio.h>