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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
40000nodes.. - 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!")