Description
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
.
Analyse
获取BST(Binary Search Tree)中居于L
与R
之间的节点之和(包括L
和R
)
BST的中序遍历即为递增的有序序列,我写的第一个版本用了中序遍历,但只是简单的判断节点的值是否处于L
,R
之间,未用到BST的性质,这样可以达到faster than 98.11%
int rangeSumBST(TreeNode* root, int L, int R) {
if (root == NULL) return 0;
int sum = 0;
inOrder(root, L, R, sum);
return sum;
}
void inOrder(TreeNode* root, int L, int R, int& sum)
{
if (root == NULL) {return;}
inOrder(root->left, L, R, sum);
if (root-> val >= L && root->val <= R)
{
sum += root->val;
}
inOrder(root->right, L, R, sum);
}
更好的做法是利用BST的性质减少一部分路径的遍历,贴上leetcode上的代码
如果遍历的当前节点的val
在L
和R
之间,递归加上该节点的两个子节点的值
当前节点的val
> L
&& val
> R
,递归加上该节点的左节点的值
当前节点的val
< L
&& val
< R
,递归加上该节点的右节点的值
int rangeSumBST(TreeNode* root, int L, int R) {
if(root == NULL)
return 0;
int sum = 0;
if(root->val >= L && root->val <= R)
return (root->val + rangeSumBST(root->left, L, R) +
rangeSumBST(root->right, L, R));
else if(root->val > L && root->val > R)
return rangeSumBST(root->left, L, R);
else if (root->val < L && root->val < R)
return rangeSumBST(root->right, L, R);
}