已知后续 中序 构建二叉树 并且层序遍历

include<bits/stdc++.h>

using namespace std;

const int N = 100;
int n;

// 定义二叉树结构体
struct Tree {
int x; // 节点的值
Tree* left; // 左子树
Tree* right; // 右子树
};

// 递归打印树的后序遍历（示例）
void pre(Tree* t) {
if (t == nullptr) return;
cout << t->x << " ";
pre(t->left);
pre(t->right);
}

// 递归打印树的中序遍历
void ino(Tree* t) {
if (t == nullptr) return;
ino(t->left);
cout << t->x << " ";
ino(t->right);
}

// 递归打印树的层序遍历
void levelOrder(Tree* root) {
if (root == nullptr) return;

queue<Tree*> q;
q.push(root);

while (!q.empty()) {
    Tree* node = q.front();
    q.pop();
    cout << node->x << " ";
    
    if (node->left) q.push(node->left);
    if (node->right) q.push(node->right);
}

}

// 从后序和中序遍历重建二叉树
Tree* buildTree(vector& postorder, vector& inorder, int& postIdx, int inStart, int inEnd) {
if (inStart > inEnd) return nullptr;

// 后序遍历的最后一个元素是根节点
int rootVal = postorder[postIdx--];

// 创建根节点
Tree* root = new Tree{rootVal, nullptr, nullptr};

// 找到根节点在中序遍历中的位置
int inIdx = -1;
for (int i = inStart; i <= inEnd; i++) {
    if (inorder[i] == rootVal) {
        inIdx = i;
        break;
    }
}

// 递归构建右子树和左子树
root->right = buildTree(postorder, inorder, postIdx, inIdx + 1, inEnd);
root->left = buildTree(postorder, inorder, postIdx, inStart, inIdx - 1);

return root;

}

int main() {
cin >> n;

vector<int> postorder(n), inorder(n);

for (int i = 0; i < n; i++) cin >> postorder[i];
for (int i = 0; i < n; i++) cin >> inorder[i];

// 初始的后序遍历索引
int postIdx = n - 1;

// 通过后序和中序遍历构建二叉树
Tree* root = buildTree(postorder, inorder, postIdx, 0, n - 1);

// 输出层序遍历
levelOrder(root);
cout << endl;

return 0;

}