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
.
Approach #1: C++. [recursive]
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int rangeSumBST(TreeNode* root, int L, int R) { ans = 0; dfs(root, L, R); return ans; } private: int ans; void dfs(TreeNode* root, int L, int R) { if (root != NULL) { if (L <= root->val && root->val <= R) { ans += root->val; } if (L < root->val) { dfs(root->left, L, R); } if (R > root->val) { dfs(root->right, L, R); } } } };
Approach #2: Java. [Iterative]
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ 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 (R > node.val) stack.push(node.right); } } return ans; } }
Approach #3: Python.
# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def rangeSumBST(self, root, L, R): """ :type root: TreeNode :type L: int :type R: int :rtype: int """ ans = 0 stack = [root] while stack: node = stack.pop() if node: if L <= node.val <= R: ans += node.val if L < node.val: stack.append(node.left) if R > node.val: stack.append(node.right) return ans