1373

1373. Maximum Sum BST in Binary Tree (Hard)

Given a binary tree root, the task is to return the maximum sum of all keys of any sub-tree which is also a Binary Search Tree (BST).

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than the node's key.
• Both the left and right subtrees must also be binary search trees.

 

Example 1:

Input: root = [1,4,3,2,4,2,5,null,null,null,null,null,null,4,6]
Output: 20
Explanation: Maximum sum in a valid Binary search tree is obtained in root node with key equal to 3.

Example 2:

Input: root = [4,3,null,1,2]
Output: 2
Explanation: Maximum sum in a valid Binary search tree is obtained in a single root node with key equal to 2.

Example 3:

Input: root = [-4,-2,-5]
Output: 0
Explanation: All values are negatives. Return an empty BST.

Example 4:

Input: root = [2,1,3]
Output: 6

Example 5:

Input: root = [5,4,8,3,null,6,3]
Output: 7

 

Constraints:

• Each tree has at most 40000 nodes..
• Each node's value is between [-4 * 10^4 , 4 * 10^4].

Related Topics:
Dynamic Programming, Binary Search Tree

Solution 1.

// OJ: https://leetcode.com/problems/maximum-sum-bst-in-binary-tree/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
    int ans = 0;
    pair<int, int> postorder(TreeNode *root) {
        pair<int, int> r{root->val, root->val};
        bool isBST = true;
        if (root->left) {
            auto left = postorder(root->left);
            if (root->val <= left.second) isBST = false;
            r.second = max(left.second, root->val);
            r.first = min(left.first, root->val);
        }
        if (root->right) {
            auto right = postorder(root->right);
            if (root->val >= right.first) isBST = false;
            r.second = max(right.second, root->val);
            r.first = min(right.first, root->val);
        }
        root->val += (root->left ? root->left->val : 0) + (root->right ? root->right->val : 0);
        if (isBST) ans = max(ans, root->val);
        return isBST ? r :  pair<int,int>{ INT_MIN, INT_MAX };
    }
public:
    int maxSumBST(TreeNode* root) {
        if (!root) return 0;
        postorder(root);
        return ans;
    }
};