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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.

• /**
* 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.offer(root);
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
int[] bstArray = {node.val, node.val, 0, node.val};
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;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def maxSumBST(self, root: Optional[TreeNode]) -> int:
def dfs(root: Optional[TreeNode]) -> tuple:
if root is None:
return 1, inf, -inf, 0
lbst, lmi, lmx, ls = dfs(root.left)
rbst, rmi, rmx, rs = dfs(root.right)
if lbst and rbst and lmx < root.val < rmi:
nonlocal ans
s = ls + rs + root.val
ans = max(ans, s)
return 1, min(lmi, root.val), max(rmx, root.val), s
return 0, 0, 0, 0

ans = 0
dfs(root)
return ans


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func maxSumBST(root *TreeNode) (ans int) {
const inf = 1 << 30
var dfs func(root *TreeNode) [4]int
dfs = func(root *TreeNode) [4]int {
if root == nil {
return [4]int{1, inf, -inf, 0}
}
l, r := dfs(root.Left), dfs(root.Right)
if l[0] == 1 && r[0] == 1 && l[2] < root.Val && root.Val < r[1] {
s := l[3] + r[3] + root.Val
ans = max(ans, s)
return [4]int{1, min(l[1], root.Val), max(r[2], root.Val), s}
}
return [4]int{}
}
dfs(root)
return
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

func min(a, b int) int {
if a < b {
return a
}
return b
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function maxSumBST(root: TreeNode | null): number {
const inf = 1 << 30;
let ans = 0;
const dfs = (root: TreeNode | null): [boolean, number, number, number] => {
if (!root) {
return [true, inf, -inf, 0];
}
const [lbst, lmi, lmx, ls] = dfs(root.left);
const [rbst, rmi, rmx, rs] = dfs(root.right);
if (lbst && rbst && lmx < root.val && root.val < rmi) {
const s = ls + rs + root.val;
ans = Math.max(ans, s);
return [true, Math.min(lmi, root.val), Math.max(rmx, root.val), s];
}
return [false, 0, 0, 0];
};
dfs(root);
return ans;
}