Note:
A subtree must include all of its descendants.
Here‘s an example:
10
/ 5 15
/ \ \
1 8 7
The Largest BST Subtree in this case is the highlighted one.
The return value is the subtree‘s size, which is 3.
Follow up:
Can you figure out ways to solve it with O(n) time complexity?
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * public int val; 5 * public TreeNode left; 6 * public TreeNode right; 7 * public TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Result 11 { 12 public int lower; 13 public int upper; 14 public int count; 15 16 public Result(int l, int u, int c) 17 { 18 lower = l; 19 upper = u; 20 count = c; 21 } 22 } 23 24 // bottom up solution, the range is calculated from bottom, we need to use a special datastructure Result to implement it 25 public class Solution { 26 public int LargestBSTSubtree(TreeNode root) { 27 if (root == null) return 0; 28 var count = new int[1]; 29 30 CountBSTNodes(root, count); 31 return count[0]; 32 } 33 34 private Result CountBSTNodes(TreeNode node, int[] count) 35 { 36 if (node == null) return new Result(Int32.MaxValue, Int32.MinValue, 0); 37 38 var left = CountBSTNodes(node.left, count); 39 var right = CountBSTNodes(node.right, count); 40 41 if (left.count < 0 || right.count < 0 || node.val <= left.upper || node.val >= right.lower) 42 { 43 return new Result(0, 0, -1); 44 } 45 46 count[0] = Math.Max(count[0], left.count + right.count + 1); 47 return new Result(Math.Min(left.lower, node.val), Math.Max(right.upper, node.val), left.count + right.count + 1); 48 } 49 } 50 51 // top down solution 52 public class Solution1 { 53 public int LargestBSTSubtree(TreeNode root) { 54 if (root == null) return 0; 55 56 int total = CountBSTNodes(root, Int32.MinValue, Int32.MaxValue); 57 if (total >= 0) return total; 58 59 return Math.Max(LargestBSTSubtree(root.left), LargestBSTSubtree(root.right)); 60 } 61 62 // returns -1 if it‘s not a BST 63 private int CountBSTNodes(TreeNode node, int min, int max) 64 { 65 if (node == null) return 0; 66 67 // quick cut 68 if (node.val <= min || node.val >= max) return -1; 69 70 int left = CountBSTNodes(node.left, min, node.val); 71 if (left < 0) 72 { 73 return -1; 74 } 75 76 int right = CountBSTNodes(node.right, node.val, max); 77 if (right < 0) 78 { 79 return -1; 80 } 81 82 return left + right + 1; 83 } 84 }
