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# 938.Range-Sum-of-BST

## 题目描述

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.

## 代码

### Approach #1 DFS

/**
* 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) {
ans = 0;
dfs(root, L, R);
return ans;
}
private void dfs(TreeNode node, int L, int R) {
if (node != null) {
if (node.val >= L && node.val <= R) {
ans += node.val;
}
// use if not else if
if (L < node.val) {
dfs(node.left, L, R);
}
if (node.val < R) {
dfs(node.right, L, R);
}
}
}
}

### Approach #2 Iterative

class Solution {
public int rangeSumBST(TreeNode root, int L, int R) {
int ans = 0;
Stack<TreeNode> stack = new Stack();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
if (node != null) {
if (L <= node.val && node.val <= R) {
ans += node.val;
}
if (L < node.val) {
stack.push(node.left);
}
if (node.val < R) {
stack.push(node.right);
}
}
}
return ans;
}
}