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RangeSumOfBST.java
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RangeSumOfBST.java
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package algorithms;
/**
* 938. Range Sum of BST
* https://leetcode.com/problems/range-sum-of-bst/
* Difficulty : Easy
* Related Topics : Tree, Recursion
*
* Given the root node of a binary search tree, return the sum of values of all nodes with value between L and R (inclusive).
*
* The binary search tree is guaranteed to have unique values.
*
*
*
* Example 1:
*
* Input: root = [10,5,15,3,7,null,18], L = 7, R = 15
* Output: 32
* Example 2:
*
* Input: root = [10,5,15,3,7,13,18,1,null,6], L = 6, R = 10
* Output: 23
*
*
* Note:
*
* The number of nodes in the tree is at most 10000.
* The final answer is guaranteed to be less than 2^31.
*
* 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;
* }
* }
*
* created by cenkc on 8/12/2020
*/
public class RangeSumOfBST {
// in order (DFS)
// Runtime: 1 ms, faster than 54.69% of Java online submissions for Range Sum of BST.
// Memory Usage: 62.3 MB, less than 5.04% of Java online submissions for Range Sum of BST.
public int rangeSumBST(TreeNode root, int L, int R) {
if (root == null) return 0;
int sum = 0;
//in order, LNR
return dfsTraversal(root, sum, L, R);
}
private int dfsTraversal(TreeNode node, int sum, int L, int R) {
if (node.left != null) sum = dfsTraversal(node.left, sum, L, R);
sum += (node != null && L <= node.val && node.val <= R ? node.val : 0);
if (node.right != null) sum = dfsTraversal(node.right, sum, L, R);
return sum;
}
/*
// level order (BFS)
// Runtime: 23 ms, faster than 5.02% of Java online submissions for Range Sum of BST.
// Memory Usage: 68.7 MB, less than 5.04% of Java online submissions for Range Sum of BST.
class Solution {
public int rangeSumBST(TreeNode root, int L, int R) {
int sum = 0;
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
while ( ! q.isEmpty()) {
TreeNode node = q.poll();
if (node != null && L <= node.val && node.val <= R) {
sum += node.val;
}
if (node.left != null) q.offer(node.left);
if (node.right != null) q.offer(node.right);
}
return sum;
}
}
*/
}