938. Range Sum of BST
Easy
Given the root
node of a binary search tree, return the sum of values of all nodes with a value in the range [low, high]
.
Example 1:
Input: root = [10,5,15,3,7,null,18], low = 7, high = 15 Output: 32
Example 2:
Input: root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10 Output: 23
Constraints:
- The number of nodes in the tree is in the range
[1, 2 * 104]
. 1 <= Node.val <= 105
1 <= low <= high <= 105
- All
Node.val
are unique.
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 | /** * 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 { int sum = 0; public int rangeSumBST(TreeNode root, int low, int high) { helper(root, low, high); return sum; } void helper(TreeNode node, int low, int high){ if(node==null) return; if(node.val>=low && node.val<=high) sum += node.val; if(!(node.val<=low)) helper(node.left, low, high); if(!(node.val>=high)) helper(node.right, low, high); } } |
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