给定二叉搜索树的根结点 root
,返回值位于范围 [low, high]
之间的所有结点的值的和。
示例 1:
输入:root = [10,5,15,3,7,null,18], low = 7, high = 15 输出:32
示例 2:
输入:root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10 输出:23
提示:
- 树中节点数目在范围
[1, 2 * 104]
内 1 <= Node.val <= 105
1 <= low <= high <= 105
- 所有
Node.val
互不相同
问题简要描述:返回位于范围 [low, high]
之间的所有结点的值的和
Java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int rangeSumBST(TreeNode root, int low, int high) {
return dfs(root, low, high);
}
int dfs(TreeNode root, int low, int high) {
if (root == null) {
return 0;
}
int x = root.val;
int ans = low <= x && x <= high ? x : 0;
if (x > low) {
ans += dfs(root.left, low, high);
}
if (x < high) {
ans += dfs(root.right, low, high);
}
return ans;
}
}
Python3
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def rangeSumBST(self, root: Optional[TreeNode], low: int, high: int) -> int:
def dfs(root):
if root is None:
return 0
x = root.val
ans = x if low <= x and x <= high else 0
if x > low:
ans += dfs(root.left)
if x < high:
ans += dfs(root.right)
return ans
return dfs(root)
TypeScript
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function rangeSumBST(root: TreeNode | null, low: number, high: number): number {
const dfs = (root) => {
if (!root) {
return 0;
}
let x = root.val;
let ans = low <= x && x <= high ? x : 0;
if (x > low) {
ans += dfs(root.left);
}
if (x < high) {
ans += dfs(root.right);
}
return ans;
}
return dfs(root);
};