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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.
题意:给定一颗二叉搜索树tree和一个整数范围[L, R],计算tree中所有节点在此范围内的值的和;
解法:最简单的方法就是遍历所有的节点,判断此节点的值是否在给定范围内即可;不过,因为给定的树是一颗搜索树,那么必定有
- root.val > root.left.val
- root.val < root.right.val
根据这个特点,我们可以做一个简单的剪枝处理
- 如果root.val < L,那么以root为根的左子树的所有节点值必然小于L;
- 如果root.val > R, 那么以root为根的右子树的所有节点值必然大于R;
Java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
int sum = 0;
public int rangeSumBST(TreeNode root, int L, int R) {
if (root == null) {
return 0;
}
if (root.val >= L && root.val <= R) {
sum += root.val;
rangeSumBST(root.left, L, R);
rangeSumBST(root.right, L, R);
} else if (root.val < L) {
rangeSumBST(root.right, L, R);
} else if (root.val > R) {
rangeSumBST(root.left, L, R);
}
return sum;
}
}