LeetCode 938. Range Sum of BST

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本文介绍了一种算法，用于解决给定二叉搜索树和两个整数L和R时，如何找到所有节点值在[L,R]范围内的节点值之和。通过递归遍历树并利用二叉搜索树的性质进行剪枝，该算法能有效地计算出结果。

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.

给定一棵二叉树、L和R，返回[L, R]中的结点的值的和。

剪枝：

//C++
class Solution {
public:
    int rangeSumBST(TreeNode* root, int L, int R) {
        if (root == NULL) return 0;
        int sum = 0;
        if (root->val > L) sum += rangeSumBST(root->left, L, R);
        if (root->val < R) sum += rangeSumBST(root->right, L, R);
        return sum + (L <= root->val && root->val <= R ? root->val : 0);
    }
};

不剪枝：

//C++
class Solution {
public:
    int rangeSumBST(TreeNode* root, int L, int R) {
        if (root == NULL) return 0;
        return rangeSumBST(root->left, L, R) + rangeSumBST(root->right, L, R) + 
            (L <= root->val && root->val <= R ? root->val : 0);
    }
};