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1373. Maximum Sum BST in Binary Tree

Description

Given a binary tree root, 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.

 

Constraints:

  • The number of nodes in the tree is in the range [1, 4 * 104].
  • -4 * 104 <= Node.val <= 4 * 104

Solutions

Solution 1: DFS

To determine whether a tree is a binary search tree, it needs to meet the following four conditions:

  • The left subtree is a binary search tree;
  • The right subtree is a binary search tree;
  • The maximum value of the left subtree is less than the value of the root node;
  • The minimum value of the right subtree is greater than the value of the root node.

Therefore, we design a function $dfs(root)$, the return value of the function is a quadruple $(bst, mi, mx, s)$, where:

  • The number $bst$ indicates whether the tree with $root$ as the root is a binary search tree. If it is a binary search tree, then $bst = 1$; otherwise $bst = 0$;
  • The number $mi$ represents the minimum value of the tree with $root$ as the root;
  • The number $mx$ represents the maximum value of the tree with $root$ as the root;
  • The number $s$ represents the sum of all nodes of the tree with $root$ as the root.

The execution logic of the function $dfs(root)$ is as follows:

If $root$ is an empty node, return $(1, +\infty, -\infty, 0)$, indicating that the empty tree is a binary search tree, the minimum value and maximum value are positive infinity and negative infinity respectively, and the sum of nodes is $0$.

Otherwise, recursively calculate the left subtree and right subtree of $root$, and get $(lbst, lmi, lmx, ls)$ and $(rbst, rmi, rmx, rs)$ respectively, then judge whether the $root$ node meets the conditions of the binary search tree.

If $lbst = 1$ and $rbst = 1$ and $lmx < root.val < rmi$, then the tree with $root$ as the root is a binary search tree, and the sum of nodes $s= ls + rs + root.val$. We update the answer $ans = \max(ans, s)$, and return $(1, \min(lmi, root.val), \max(rmx, root.val), s)$.

Otherwise, the tree with $root$ as the root is not a binary search tree, we return $(0, 0, 0, 0)$.

We call $dfs(root)$ in the main function. After execution, the answer is $ans$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the number of nodes in the binary tree.

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# 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
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int ans;
    private final int inf = 1 << 30;

    public int maxSumBST(TreeNode root) {
        dfs(root);
        return ans;
    }

    private int[] dfs(TreeNode root) {
        if (root == null) {
            return new int[] {1, inf, -inf, 0};
        }
        var l = dfs(root.left);
        var r = dfs(root.right);
        int v = root.val;
        if (l[0] == 1 && r[0] == 1 && l[2] < v && r[1] > v) {
            int s = v + l[3] + r[3];
            ans = Math.max(ans, s);
            return new int[] {1, Math.min(l[1], v), Math.max(r[2], v), s};
        }
        return new int[4];
    }
}
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxSumBST(TreeNode* root) {
        int ans = 0;
        const int