forked from fishercoder1534/Leetcode
-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy path_1373.java
72 lines (69 loc) · 2.59 KB
/
_1373.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
package com.fishercoder.solutions;
import com.fishercoder.common.classes.TreeNode;
/**
* 1373. Maximum Sum BST in Binary Tree
*
* Given a binary tree root, the task is to 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.
*
* Example 4:
* Input: root = [2,1,3]
* Output: 6
*
* Example 5:
* Input: root = [5,4,8,3,null,6,3]
* Output: 7
*
* Constraints:
* Each tree has at most 40000 nodes..
* Each node's value is between [-4 * 10^4 , 4 * 10^4].
* */
public class _1373 {
public static class Solution1 {
/**
* credit: https://leetcode.com/problems/maximum-sum-bst-in-binary-tree/discuss/532021/Java-Post-Order
* */
public int maxSumBST(TreeNode root) {
return postOrder(root)[4];
}
/**
* result[0] means this tree is a BST
* result[1] means the sum of this tree
* result[2] means the left boundary
* result[3] means the right boundary
* result[4] means the global max sum
* */
private int[] postOrder(TreeNode root) {
if (root == null) {
return new int[]{1, 0, Integer.MAX_VALUE, Integer.MIN_VALUE, 0};
}
int[] leftSide = postOrder(root.left);
int[] rightSide = postOrder(root.right);
int localMax = Math.max(leftSide[4], rightSide[4]);
if (leftSide[0] == 1 && rightSide[0] == 1 && root.val > leftSide[3] && root.val < rightSide[2]) {
int sum = root.val + leftSide[1] + rightSide[1];
return new int[]{1, sum, leftSide[2] == Integer.MAX_VALUE ? root.val : leftSide[2], rightSide[3] == Integer.MIN_VALUE ? root.val : rightSide[3], Math.max(localMax, sum)};
} else {
return new int[]{0, 0, 0, 0, localMax};
}
}
}
}