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333. Largest BST Subtree πŸ”’

Description

Given the root of a binary tree, find the largest subtree, which is also a Binary Search Tree (BST), where the largest means subtree has the largest number of nodes.

A Binary Search Tree (BST) is a tree in which all the nodes follow the below-mentioned properties:

  • The left subtree values are less than the value of their parent (root) node's value.
  • The right subtree values are greater than the value of their parent (root) node's value.

Note: A subtree must include all of its descendants.

 

Example 1:

Input: root = [10,5,15,1,8,null,7]
Output: 3
Explanation: The Largest BST Subtree in this case is the highlighted one. The return value is the subtree's size, which is 3.

Example 2:

Input: root = [4,2,7,2,3,5,null,2,null,null,null,null,null,1]
Output: 2

 

Constraints:

  • The number of nodes in the tree is in the range [0, 104].
  • -104 <= Node.val <= 104

 

Follow up: Can you figure out ways to solve it with O(n) time complexity?

Solutions

Solution 1

Thinking

Find the size of the largest BST subtree. Validating a BST at every node repeats work. A subtree is a BST iff both sides are BSTs and the root sits between their extrema.

Postorder returns \((\min,\max,size)\). If left-max \(<\) root \(<\) right-min, merge sizes and update the answer; otherwise return sentinels \((-\infty,\infty,0)\) so ancestors cannot absorb it. An empty tree returns \((\infty,-\infty,0)\) so a leaf succeeds.

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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 largestBSTSubtree(self, root: Optional[TreeNode]) -> int:
        def dfs(root):
            if root is None:
                return inf, -inf, 0
            lmi, lmx, ln = dfs(root.left)
            rmi, rmx, rn = dfs(root.right)
            nonlocal ans
            if lmx < root.val < rmi:
                ans = max(ans, ln + rn + 1)
                return min(lmi, root.val), max(rmx, root.val), ln + rn + 1
            return -inf, inf, 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;

    public int largestBSTSubtree(TreeNode root) {
        ans = 0;
        dfs(root);
        return ans;
    }

    private int[] dfs(TreeNode root) {
        if (root == null) {
            return new int[] {Integer.MAX_VALUE, Integer.MIN_VALUE, 0};
        }
        int[] left = dfs(root.left);
        int[] right = dfs(root.right);
        if (left[1] < root.val && root.val < right[0]) {
            ans = Math.max(ans, left[2] + right[2] + 1);
            return new int[] {
                Math.min(root.val, left[0]), Math.max(root.val, right[1]), left[2] + right[2] + 1};
        }
        return new int[] {Integer.MIN_VALUE, Integer.MAX_VALUE, 0};
    }
}

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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 ans;

    int largestBSTSubtree(TreeNode* root) {
        ans = 0;
        dfs(root);
        return ans;
    }

    vector<int> dfs(TreeNode* root) {
        if (!root) return {INT_MAX, INT_MIN, 0};
        auto left = dfs(root->left);
        auto right = dfs(root->right);
        if (left[1] < root->val && root->val < right[0]) {
            ans = max(ans, left[2] + right[2] + 1);
            return {min(root->val, left[0]), max(root->val, right[1]), left[2] + right[2] + 1};
        }
        return {INT_MIN, INT_MAX, 0};
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func largestBSTSubtree(root *TreeNode) int {
    ans := 0
    var dfs func(root *TreeNode) []int
    dfs = func(root *TreeNode) []int {
        if root == nil {
            return []int{math.MaxInt32, math.MinInt32, 0}
        }
        left := dfs(root.Left)
        right := dfs(root.Right)
        if left[1] < root.Val && root.Val < right[0] {
            ans = max(ans, left[2]+right[2]+1)
            return []int{min(root.Val, left[0]), max(root.Val, right[1]), left[2] + right[2] + 1}
        }
        return []int{math.MinInt32, math.MaxInt32, 0}
    }
    dfs(root)
    return ans
}

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