1119. 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.

71 2 3 4 6 7 52 6 7 4 5 3 1Sample Output 1:
Yes2 1 6 4 7 3 5Sample Input 2:
41 2 3 42 4 3 1Sample Output 2:
No2 1 3 4
#include <cstdio>#include <vector>using namespace std;vector<int> ans;int *pre, *post, unique = 1;int findFromPre (int x, int l, int r) {	for (int i = l; i <= r; i++) {		if (x == pre[i]) {			return i;		}	}	return -1;}void setIn (int prel, int prer, int postl, int postr) {	if (prel == prer) {		ans.push_back(pre[prel]);		return;	}	if (pre[prel] == post[postr]) {		int x = findFromPre(post[postr - 1], prel + 1, prer);		if (x - prel > 1) {			setIn(prel + 1, x - 1, postl, postl + x - prel - 2);			ans.push_back(post[postr]);			setIn(x, prer, postl + x - prel - 2 + 1, postr - 1);		} else {			unique = 0;			ans.push_back(post[postr]);			setIn(x, prer, postl + x - prel - 2 + 1, postr - 1);		}	} }int main() {	int n = 0;	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]);	}	setIn(0, n - 1, 0, n - 1);	printf("%s/n", unique ? "Yes" : "No");	printf("%d", ans[0]);	for (int i = 1; i < ans.size(); i++) {		printf(" %d", ans[i]);	}	printf("/n");	return 0;}
#include <iostream>  using namespace std;const int maxn = 31;int n, index = 0;int pre[maxn], post[maxn];bool flag = true;struct Node {	int data;	Node *lchild, *rchild;} *root;Node *create(int preL, int preR, int postL, int postR){	if (preL > preR) return NULL;//不合理就返回NULL	Node *node = new Node;//建立根結點	node->data = pre[preL];	node->lchild = NULL;	node->rchild = NULL;	if (preL == preR)		return node;//若只有一個節點那么直接返回即可	int k = 0;	for (k = preL + 1; k <= preR; k++)	{		if (pre[k] == post[postR - 1]) break;//從前序中找后序中子樹根的位置	}	if (k - preL > 1)	{		node->lchild = create(preL + 1, k - 1, postL, postL + k - preL - 2);		node->rchild = create(k, preR, postL + k - preL - 1, postR - 1);	}	else	{		flag = false;		node->rchild = create(k, preR, postL + k - preL - 1, postR - 1);	}	return node;}void inOrder(Node *node){	if (node == NULL) return;	inOrder(node->lchild);	if (index < n - 1)		cout << node->data << " ";	else cout << node->data << endl;	index++;	inOrder(node->rchild);}int main(){	cin >> n;	for (int i = 0; i < n; ++i) cin >> pre[i];	for (int i = 0; i < n; ++i) cin >> post[i];	root = create(0, n - 1, 0, n - 1);	if (flag) cout << "Yes/n";	else cout << "No/n";	inOrder(root);	return 0;}/*這題我開始做有點難以理解，后來想清楚了：我解釋下【1】 2 3 4 6 7 5 2 6 7 4 5 3 【1】 一定要記住，前序是根 左 右的 遍歷順序而后序是左 右 根的遍 歷順序那么通過確定后序中3在前 序中的位置，那么就知道如何劃分 子樹了，因為3一定是1的右子樹的 根，那么在前序中，3之前1之后 的序列就是左子樹，3到5之間都 為右子樹的一部分，再遞歸處理即可 但是若根結點只有一顆子樹，那么 得確定到底是左還是右，自己決定 這也是前+后不一定能唯一確定樹的原因 1 2 3 4 2 4 3 1 1為根，找右子樹的根即3在前序中的位置 那么2是左子樹，3 4是右子樹，繼續遞歸 2只有一個直接返回，3 4中3是根，4是子 樹，但是左還是右？自己決定，這種情況就是不能唯一確定*/