算法——Week5

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

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解题思路
这道题就是遍历出在L和R之间的节点值之和。我用深度优先搜索遍历这个二分搜索树，满足条件的节点值加到结果上。

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代码如下

/**
 * 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 ans = 0;
    int rangeSumBST(TreeNode* root, int L, int R) {
        dfs(root, L, R);
        return ans;
    }
    void dfs(TreeNode *root, int L, int R) {
        if(root != NULL) {
            if(root->val >= L && root->val <= R) {
                ans = ans + root->val;                
            }
            if(root->val > L) {
                dfs(root->left, L, R);
            }
            if(root->val < R) {
                dfs(root->right, L, R);
            }
        }
    }
};