Formatted question description: https://leetcode.ca/all/1373.html

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/

// 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;
    }
};

Java

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public int maxSumBST(TreeNode root) {
            if (root == null)
                return 0;
            if (root.left == null && root.right == null)
                return root.val;
            List<TreeNode> list = new ArrayList<TreeNode>();
            Map<TreeNode, int[]> map = new HashMap<TreeNode, int[]>();
            Queue<TreeNode> queue = new LinkedList<TreeNode>();
            queue.offer(root);
            while (!queue.isEmpty()) {
                TreeNode node = queue.poll();
                int[] bstArray = {node.val, node.val, 0, node.val};
                list.add(node);
                map.put(node, bstArray);
                TreeNode left = node.left, right = node.right;
                if (left != null)
                    queue.offer(left);
                if (right != null)
                    queue.offer(right);
            }
            int maxSum = 0;
            for (int i = list.size() - 1; i >= 0; i--) {
                TreeNode node = list.get(i);
                int[] bstArray = map.getOrDefault(node, new int[]{node.val, node.val, 0, 0});
                TreeNode left = node.left, right = node.right;
                if (left == null && right == null) {
                    bstArray[2] = 1;
                    bstArray[3] = node.val;
                    map.put(node, bstArray);
                    maxSum = Math.max(maxSum, bstArray[3]);
                } else if (right == null) {
                    int[] leftBstArray = map.getOrDefault(left, new int[]{left.val, left.val, 0, 0});
                    if (leftBstArray[2] == 1 && leftBstArray[1] < node.val) {
                        bstArray[0] = leftBstArray[0];
                        bstArray[2] = 1;
                        bstArray[3] = leftBstArray[3] + node.val;
                        map.put(node, bstArray);
                        maxSum = Math.max(maxSum, bstArray[3]);
                    }
                } else if (left == null) {
                    int[] rightBstArray = map.getOrDefault(right, new int[]{right.val, right.val, 0, 0});
                    if (rightBstArray[2] == 1 && rightBstArray[0] > node.val) {
                        bstArray[1] = rightBstArray[1];
                        bstArray[2] = 1;
                        bstArray[3] = rightBstArray[3] + node.val;
                        map.put(node, bstArray);
                        maxSum = Math.max(maxSum, bstArray[3]);
                    }
                } else {
                    int[] leftBstArray = map.getOrDefault(left, new int[]{left.val, left.val, 0, 0});
                    int[] rightBstArray = map.getOrDefault(right, new int[]{right.val, right.val, 0, 0});
                    if (leftBstArray[2] == 1 && rightBstArray[2] == 1 && leftBstArray[1] < node.val && rightBstArray[0] > node.val) {
                        bstArray[0] = leftBstArray[0];
                        bstArray[1] = rightBstArray[1];
                        bstArray[2] = 1;
                        bstArray[3] = leftBstArray[3] + rightBstArray[3] + node.val;
                        map.put(node, bstArray);
                        maxSum = Math.max(maxSum, bstArray[3]);
                    }
                }
            }
            return maxSum;
        }
    }
    
  • // OJ: https://leetcode.com/problems/maximum-sum-bst-in-binary-tree/
    // 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;
        }
    };
    
  • print("Todo!")
    

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