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

Example 2:

Input: root = [10,5,15,3,7,13,18,1,null,6], L = 6, R = 10
Output: 23

Note:

1. The number of nodes in the tree is at most 10000.
2. 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