class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 求二叉树的右视图
     * @param preOrder int整型vector 先序遍历
     * @param inOrder int整型vector 中序遍历
     * @return int整型vector
     */
    TreeNode* build(
    const vector<int>& pre, int preL, int preR, 
    const vector<int>& in, int inL, int inR)
{
    if (preL > preR || inL > inR) return nullptr;

    TreeNode* root = new TreeNode(pre[preL]);

    int Search = find(in.begin() + inL, in.begin() + inR + 1, pre[preL]) - in.begin();
    int k = Search - inL;  // 左子树节点数

    root->left = build(
        pre,
        preL + 1,
        preL + k,
        in,
        inL,
        Search - 1
    );

    root->right = build(
        pre,
        preL + k + 1,
        preR,
        in,
        Search + 1,
        inR
    );

    return root;
}

    vector<int> solve(vector<int>& preOrder, vector<int>& inOrder) {
    TreeNode* root = build(
        preOrder, 0, preOrder.size() - 1,
        inOrder, 0, inOrder.size() - 1
    );

    vector<int> result;
    if (!root) return result;

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

    while (!q.empty()) {
        int sz = q.size();   // 当前层节点数
        for (int i = 0; i < sz; i++) {
            TreeNode* t = q.front(); q.pop();

            // 当前层最后一个节点
            if (i == sz - 1) {
                result.push_back(t->val);
            }

            if (t->left) q.push(t->left);
            if (t->right) q.push(t->right);
        }
    }
    return result;
}

};