654. Maximum Binary Tree

You are given an integer array nums with no duplicates. A maximum binary tree can be built recursively from nums using the following algorithm:

• Create a root node whose value is the maximum value in nums.
• Recursively build the left subtree on the subarray prefix to the left of the maximum value.
• Recursively build the right subtree on the subarray suffix to the right of the maximum value.

Return the maximum binary tree built from nums.

• divide and conquer solution

/**
 * 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 {
    TreeNode* dnc(vector<int>& A, int l, int r) {
        if(l > r) return nullptr;
        int m = max_element(begin(A) + l, begin(A) + r + 1) - begin(A);
        TreeNode* res = new TreeNode(A[m]);
        res->left = dnc(A, l, m - 1);
        res->right = dnc(A, m + 1, r);
        return res;
    }
public:
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        return dnc(nums, 0, nums.size() - 1);
    }
};

• monotonic stack solution

class Solution {
public:
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        vector<TreeNode*> st;
        
        for(auto& n : nums) {
            TreeNode* node = new TreeNode(n);
            while(!st.empty() and st.back()->val < n) {
                node->left = st.back();
                st.pop_back();
            }
            if(!st.empty())
                st.back()->right = node;
            st.push_back(node);
        }
        
        return st[0];
    }
};