Welcome to Subscribe On Youtube
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
- The number
indicates whether the tree withbst as the root is a binary search tree. If it is a binary search tree, thenroot ; otherwisebst=1 ;bst=0 - The number
represents the minimum value of the tree withmi as the root;root - The number
represents the maximum value of the tree withmx as the root;root - The number
represents the sum of all nodes of the tree withs as the root.root
The execution logic of the function
If
Otherwise, recursively calculate the left subtree and right subtree of
If
Otherwise, the tree with
We call
The time complexity is
-
/** * 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; }