Leetcode: 938
938. Range Sum of BST
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
( 7, 10, 15) => the value between 7~ 15
Solutions:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
int ans;
public int rangeSumBST(TreeNode root, int L, int R) {
// //iterative
// int count = 0;
// Stack<TreeNode> a = new Stack<>();
// a.add(root);
// while(a.size() > 0){
// TreeNode temp = a.pop();
// if(temp == null){
// continue;
// }
// if( temp.val > L) a.push( temp.left );
// if( temp.val < R) a.push( temp.right );
// if( L <= temp.val && temp.val <= R) count += temp.val;
// }
// return count;
//recursive
if( root == null) return 0;
if( root.val < L) return rangeSumBST( root.right, L, R);
if( root.val > R) return rangeSumBST( root.left, L, R);
//consider from this line, we need add the value together, so we need check the situation
//up three lines give us method to do in different situations
//so those are false situation
return root.val + rangeSumBST( root.right, L, R) + rangeSumBST( root.left, L, R);
}
}
Offical solution:
class Solution {
int ans;
public int rangeSumBST(TreeNode root, int L, int R) {
ans = 0;
dfs(root, L, R);
return ans;
}
public void dfs(TreeNode node, int L, int R) {
if (node != null) {
if (L <= node.val && node.val <= R)
ans += node.val;
if (L < node.val)
dfs(node.left, L, R);
if (node.val < R)
dfs(node.right, L, R);
}
}
}