938. Range Sum of BST

  • Hardness: \(\color{green}\textsf{Easy}\)
  • Ralated Topics: Tree、Depth-First Search、Binary Search Tree、Binary Tree

一、題目

Given the root node of a binary search tree and two integers low and high, return the sum of values of all nodes with a value in the inclusive range [low, high].

Example 1:
bst1

  • Input: root = [10,5,15,3,7,null,18], low = 7, high = 15
  • Output: 32
  • Explanation: Nodes 7, 10, and 15 are in the range [7, 15]. 7 + 10 + 15 = 32.

Example 2: bst2

  • Input: root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10
  • Output: 23
  • Explanation: Nodes 6, 7, and 10 are in the range [6, 10]. 6 + 7 + 10 = 23.

Constraints:

  • The number of nodes in the tree is in the range [1, 2 * 104].
  • 1 <= Node.val <= 105
  • 1 <= low <= high <= 105
  • All Node.val are unique.

二、分析

  • 最簡單的方法可以遍歷過一次，將符合條件的數值加起來，可得到解。
  • 進一步可以思考，可以怎麼樣做到剪枝(pruning)。
    • 當 num < low 時，只有其右子葉有可能有符合條件的子葉。
    • 當 num > high 時，只有其左子葉有可能有符合條件的子葉。

三、解題

1. DFS

  • Time complexity: \(O(n)\)
  • Space complexity: \(O(1)\)
int rangeSumBST(TreeNode* root, int low, int high) {
    if (!root) return 0;
    int sum = root->val <= high && root->val >= low ? root->val : 0;
    return sum + rangeSumBST(root->left, low, high) + rangeSumBST(root->right, low, high);
}

2. Binary Search Tree

  • Time complexity: \(O(n)\)
  • Space complexity: \(O(1)\)
int rangeSumBST(TreeNode* root, int low, int high) {
    if (!root) return 0;
    int val = root->val >= low && root->val <= high ? root->val : 0;
    int left = root->val < low ? 0 : rangeSumBST(root->left, low, high);
    int right = root->val > high ? 0 : rangeSumBST(root->right, low, high);
    return left + right + val;
}

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