PAT 1119 Pre- and Post-order Traversals——搞清楚前序和后序生成树是不确定的！

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本文详细解析了根据前序和后序遍历构建二叉树的过程，并通过C++代码实现了这一算法。讨论了在叶节点处可能出现的构建不确定性，强调了构建树并遍历以检查正确性的必要性。

分析

前序：根左右

后序：左右根

建树的核心在于理清楚左右各部分有多少

yes的情形：在叶结点，某一块3个时，此时就是唯一情况

no的情形：在叶结点，某一块2个时，无法判断是左边还是右边，此时就会出现2种情况

代码

#include <bits/stdc++.h>
using namespace std;

int quantity, x;
bool flag{ true }, first{ true };
vector<int>preorder, postorder;

struct Tree {
	int key;
	Tree* lchild, * rchild;
	Tree(int x) {
		key = x;
		lchild = rchild = NULL;
	}
};

void InPut();
Tree* BuildTree(Tree*, int, int, int, int);
void WalkPrint(Tree*);
int main() {
	InPut();
	Tree* root = NULL;
	root = BuildTree(root, 0, quantity - 1, 0, quantity - 1);
	flag ? printf("Yes\n") : printf("No\n");
	WalkPrint(root);
	printf("\n");
	return 0;
}
void InPut() {
	scanf("%d", &quantity);
	preorder.resize(quantity);
	postorder.resize(quantity);
	for (int i = 0; i < quantity; i++) {
		scanf("%d", &x);
		preorder[i] = x;
	}
	for (int i = 0; i < quantity; i++) {
		scanf("%d", &x);
		postorder[i] = x;
	}
}
Tree* BuildTree(Tree* T, int pre_begin, int pre_end, int post_begin, int post_end) {
	if (pre_end >= pre_begin ) {
		T = new Tree(preorder[pre_begin]);
		if (post_end - 1 >= 0) {
			int w = find(preorder.begin(), preorder.end(), postorder[post_end - 1]) - preorder.begin();
			int p = w - pre_begin - 1;
			if (p >= 0) {
				T->lchild = BuildTree(T->lchild, pre_begin + 1, pre_begin + p, post_begin, post_begin + p - 1);
				T->rchild = BuildTree(T->rchild, pre_begin + p + 1, pre_end, post_begin + p, post_end - 1);
			}
		}
	}
	if (pre_end - pre_begin == 1) {
		flag = false;
	}
	return T;
}
void WalkPrint(Tree* T) {
	if (T == NULL)
		return;
	WalkPrint(T->lchild);
	first ? printf("%d", T->key), first = false : printf(" %d", T->key);
	WalkPrint(T->rchild);
}