938.Range Sum of BST
===

###### tags: `Easy`,`Tree`,`Binary Tree`,`DFS`,`Binary Search Tree`

[938. Range Sum of BST](https://leetcode.com/problems/range-sum-of-bst/)

### 題目描述

Given the `root` node of a binary search tree and two integers `low` and `high`, return *the sum of values of all nodes with a value in the **inclusive** range*`[low, high]`.

### 範例

**Example 1:**


Input: root = [10,5,15,3,7,null,18], low = 7, high = 15
Output: 32
Explanation: Nodes 7, 10, and 15 are in the range [7, 15]. 7 + 10 + 15 = 32.

**Example 2:**


Input: root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10
Output: 23
Explanation: Nodes 6, 7, and 10 are in the range [6, 10]. 6 + 7 + 10 = 23.

**Constraints**:

* The number of nodes in the tree is in the range [1, 2 * 10^4^].
* 1 <= `Node.val` <= 10^5^
* 1 <= `low` <= `high` <= 10^5^
* All `Node.val` are **unique**.

### 解答

#### Javascript

function rangeSumBST(root, low, high) {
  if (root === null) return 0;
  if (root.val < low) return rangeSumBST(root.right, low, high);
  if (root.val > high) return rangeSumBST(root.left, low, high);
  
  return root.val + rangeSumBST(root.left, low, high) + rangeSumBST(root.right, low, high);
}

function rangeSumBST(root, low, high) {
  let sum = 0;
  const stack = [root];
  while (stack.length > 0) {
    const node = stack.pop();
    if (node === null) continue;
    if (node.val >= low && node.val <= high) sum += node.val;
    if (node.val > low) stack.push(node.left);
    if (node.val < high) stack.push(node.right);
  }

  return sum;
}

> 多寫一個非遞迴的版本
> [name=Marsgoat][time= Dec 7, 2022]

#### Python
##### Recursion

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rangeSumBST(self, root: Optional[TreeNode], low: int, high: int) -> int:
        if not root: return 0
        elif low <= root.val <= high:
            return root.val + self.rangeSumBST(root.left, low, high) + self.rangeSumBST(root.right, low, high)
        else:
            return self.rangeSumBST(root.left, low, high) + self.rangeSumBST(root.right, low, high)

##### Iterative

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rangeSumBST(self, root: Optional[TreeNode], low: int, high: int) -> int:
        stack = [root]
        res = 0

        while stack:
            node = stack.pop()
            if low <= node.val <= high:
                res += node.val
            
            if node.left:
                stack.append(node.left)
            if node.right:
                stack.append(node.right)

        return res

> [name=Kobe][time= Dec 7, 2022]

#### C#

public class Solution {
    public int RangeSumBST(TreeNode root, int low, int high) {
        if (root == null) return 0;

        if (root.val < low) {
            return RangeSumBST(root.right, low, high);
        }

        if (root.val > high) {
            return RangeSumBST(root.left, low, high);
        }
        
        return root.val + RangeSumBST(root.left, low, high) + RangeSumBST(root.right, low, high);
    }
}

> [name=Jim][time= Dec 7, 2022]

### Reference

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