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<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;
    }
};

# 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;
}

1373 - Maximum Sum BST in Binary Tree

1373. Maximum Sum BST in Binary Tree (Hard)

Solution 1.

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1373. Maximum Sum BST in Binary Tree (Hard)

Solution 1.

All Problems

All Solutions