本文共 1764 字，大约阅读时间需要 5 分钟。

题目：

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 2³¹.

解释：

# Definition for a binary tree node.# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution:    def rangeSumBST(self, root, L, R):        """        :type root: TreeNode        :type L: int        :type R: int        :rtype: int        """        self.sum=0        def mid(root):            if root.left:                mid(root.left)            if root.val>=L and root.val<=R:                self.sum+=root.val            if root.right:                mid(root.right)        if root:            mid(root)        return self.sum

c++代码：

/** * 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 global_sum=0;    int global_L;    int global_R;    int rangeSumBST(TreeNode* root, int L, int R) {
           global_L=L;        global_R=R;        if (root)            mid(root);        return global_sum;    }    void mid(TreeNode* root)    {
           if (root->left)            mid(root->left);        if(root->val>=global_L && root->val<=global_R)            global_sum+=root->val;        if(root->right)            mid(root->right);    } };

总结：

转载地址：http://srlcn.baihongyu.com/