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 $d f s (r o o t) <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi><mi>f</mi><mi>s</mi><mo stretchy="false">(</mo><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi><mo stretchy="false">)</mo></math>$ , the return value of the function is a quadruple $(b s t, m i, m x, s) <math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">(</mo><mi>b</mi><mi>s</mi><mi>t</mi><mo>,</mo><mi>m</mi><mi>i</mi><mo>,</mo><mi>m</mi><mi>x</mi><mo>,</mo><mi>s</mi><mo stretchy="false">)</mo></math>$ , where:

The number $b s t <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>b</mi><mi>s</mi><mi>t</mi></math>$ indicates whether the tree with $r o o t <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi></math>$ as the root is a binary search tree. If it is a binary search tree, then $b s t = 1 <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>b</mi><mi>s</mi><mi>t</mi><mo>=</mo><mn>1</mn></math>$ ; otherwise $b s t = 0 <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>b</mi><mi>s</mi><mi>t</mi><mo>=</mo><mn>0</mn></math>$ ;
The number $m i <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mi>i</mi></math>$ represents the minimum value of the tree with $r o o t <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi></math>$ as the root;
The number $m x <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mi>x</mi></math>$ represents the maximum value of the tree with $r o o t <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi></math>$ as the root;
The number $s <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>s</mi></math>$ represents the sum of all nodes of the tree with $r o o t <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi></math>$ as the root.

The execution logic of the function $d f s (r o o t) <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi><mi>f</mi><mi>s</mi><mo stretchy="false">(</mo><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi><mo stretchy="false">)</mo></math>$ is as follows:

If $r o o t <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi></math>$ is an empty node, return $(1, + \infty, - \infty, 0) <math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">(</mo><mn>1</mn><mo>,</mo><mo>+</mo><mi mathvariant="normal">\infty</mi><mo>,</mo><mo>-</mo><mi mathvariant="normal">\infty</mi><mo>,</mo><mn>0</mn><mo stretchy="false">)</mo></math>$ , 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 $0 <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0</mn></math>$ .

Otherwise, recursively calculate the left subtree and right subtree of $r o o t <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi></math>$ , and get $(l b s t, l m i, l m x, l s) <math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">(</mo><mi>l</mi><mi>b</mi><mi>s</mi><mi>t</mi><mo>,</mo><mi>l</mi><mi>m</mi><mi>i</mi><mo>,</mo><mi>l</mi><mi>m</mi><mi>x</mi><mo>,</mo><mi>l</mi><mi>s</mi><mo stretchy="false">)</mo></math>$ and $(r b s t, r m i, r m x, r s) <math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">(</mo><mi>r</mi><mi>b</mi><mi>s</mi><mi>t</mi><mo>,</mo><mi>r</mi><mi>m</mi><mi>i</mi><mo>,</mo><mi>r</mi><mi>m</mi><mi>x</mi><mo>,</mo><mi>r</mi><mi>s</mi><mo stretchy="false">)</mo></math>$ respectively, then judge whether the $r o o t <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi></math>$ node meets the conditions of the binary search tree.

If $l b s t = 1 <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>l</mi><mi>b</mi><mi>s</mi><mi>t</mi><mo>=</mo><mn>1</mn></math>$ and $r b s t = 1 <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>r</mi><mi>b</mi><mi>s</mi><mi>t</mi><mo>=</mo><mn>1</mn></math>$ and $l m x < r o o t . v a l < r m i <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>l</mi><mi>m</mi><mi>x</mi><mo><</mo><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi><mo>.</mo><mi>v</mi><mi>a</mi><mi>l</mi><mo><</mo><mi>r</mi><mi>m</mi><mi>i</mi></math>$ , then the tree with $r o o t <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi></math>$ as the root is a binary search tree, and the sum of nodes $s = l s + r s + r o o t . v a l <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>s</mi><mo>=</mo><mi>l</mi><mi>s</mi><mo>+</mo><mi>r</mi><mi>s</mi><mo>+</mo><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi><mo>.</mo><mi>v</mi><mi>a</mi><mi>l</mi></math>$ . We update the answer $a n s = max (a n s, s) <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi><mi>n</mi><mi>s</mi><mo>=</mo><mo data-mjx-texclass="OP" movablelimits="true">max</mo><mo stretchy="false">(</mo><mi>a</mi><mi>n</mi><mi>s</mi><mo>,</mo><mi>s</mi><mo stretchy="false">)</mo></math>$ , and return $(1, min (l m i, r o o t . v a l), max (r m x, r o o t . v a l), s) <math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">(</mo><mn>1</mn><mo>,</mo><mo data-mjx-texclass="OP" movablelimits="true">min</mo><mo stretchy="false">(</mo><mi>l</mi><mi>m</mi><mi>i</mi><mo>,</mo><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi><mo>.</mo><mi>v</mi><mi>a</mi><mi>l</mi><mo stretchy="false">)</mo><mo>,</mo><mo data-mjx-texclass="OP" movablelimits="true">max</mo><mo stretchy="false">(</mo><mi>r</mi><mi>m</mi><mi>x</mi><mo>,</mo><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi><mo>.</mo><mi>v</mi><mi>a</mi><mi>l</mi><mo stretchy="false">)</mo><mo>,</mo><mi>s</mi><mo stretchy="false">)</mo></math>$ .

Otherwise, the tree with $r o o t <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi></math>$ as the root is not a binary search tree, we return $(0, 0, 0, 0) <math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">(</mo><mn>0</mn><mo>,</mo><mn>0</mn><mo>,</mo><mn>0</mn><mo>,</mo><mn>0</mn><mo stretchy="false">)</mo></math>$ .

We call $d f s (r o o t) <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi><mi>f</mi><mi>s</mi><mo stretchy="false">(</mo><mi>r</mi><mi>o</mi><mi>o</mi><mi>t</mi><mo stretchy="false">)</mo></math>$ in the main function. After execution, the answer is $a n s <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi><mi>n</mi><mi>s</mi></math>$ .

The time complexity is $O (n) <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></math>$ , and the space complexity is $O (n) <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></math>$ . Where $n <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>n</mi></math>$ is the number of nodes in the binary tree.

/**
 * 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];
    }
}

/**
 * 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 inf = 1 << 30;

        function<vector<int>(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return vector<int>{1, inf, -inf, 0};
            }
            auto l = dfs(root->left);
            auto r = dfs(root->right);
            int v = root->val;
            if (l[0] && r[0] && l[2] < v && v < r[1]) {
                int s = l[3] + r[3] + v;
                ans = max(ans, s);
                return vector<int>{1, min(l[1], v), max(r[2], v), s};
            }
            return vector<int>(4);
        };
        dfs(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
}

/**
 * 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

Description

Solutions

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1373. Maximum Sum BST in Binary Tree

Description

Solutions

All Problems

All Solutions