998. Maximum Binary Tree II

A maximum tree is a tree where every node has a value greater than any other value in its subtree.

You are given the root of a maximum binary tree and an integer val.

Just as in the previous problem, the given tree was constructed from a list a (root = Construct(a)) recursively with the following Construct(a) routine:

• If a is empty, return null.
• Otherwise, let a[i] be the largest element of a. Create a root node with the value a[i].
• The left child of root will be Construct([a[0], a[1], …, a[i - 1]]).
• The right child of root will be Construct([a[i + 1], a[i + 2], …, a[a.length - 1]]).
• Return root.

Note that we were not given a directly, only a root node root = Construct(a).

Suppose b is a copy of a with the value val appended to it. It is guaranteed that b has unique values.

Return Construct(b).

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
    vector<int> helper(TreeNode* node) {
        if(!node) return {};
        auto res = helper(node->left);
        res.push_back(node->val);
        for(auto& r : helper(node->right))
            res.push_back(r);
        return res;
    }
    TreeNode* helper(vector<int>& A, int l, int r) {
        if(l >= r) return nullptr;
        int ma = max_element(begin(A) + l, begin(A) + r) - begin(A);
        TreeNode* node = new TreeNode(A[ma]);
        node->left = helper(A, l, ma);
        node->right = helper(A, ma + 1, r);
        return node;
    }
public:
    TreeNode* insertIntoMaxTree(TreeNode* root, int val) {
        auto A = helper(root);
        A.push_back(val);
        return helper(A, 0, A.size());
    }
};