Last Updated on 2022-12-07 by Clay
題目
Given therootnode of a binary search tree and two integerslowandhigh, return the sum of values of all nodes with a value in the inclusive range[low, high].
Constraints:
- The number of nodes in the tree is in the range
[1, 2 * 104]. 1 <= Node.val <= 1051 <= low <= high <= 105- All
Node.valare unique.
題目給定兩個整數 low 以及 high，而我們要做的就是把二元搜索樹中在 [low, high] 範圍中的所有節點數值加總並返回。
解題思路
這題目第一感想到的就是使用遞迴去解，並隨時判斷是否該將子節點放入遞迴函式中。比方說，如果當前節點的數值已經小於 low，那麼便沒有必要將當前節點的左節點繼續進行遞迴...... 這也是一種剪枝的動作，稍微提升一些速度。
=== (2022-12-07 更新) ===
回頭看一年前寫的程式碼，覺得可讀性不怎麼樣。所以 C++ 的部份重新修改過了，感覺這樣比較好懂。
C++ 範例程式碼
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int rangeSumBST(TreeNode* root, int low, int high) {
// Base case
if (!root) {
return 0;
}
// Init
int rangeSum = 0;
// If the value of root is inclusive [low, high].
if (root->val >= low && root->val <= high) {
rangeSum += root->val;
}
// Check the left node and right node
rangeSum += rangeSumBST(root->left, low, high);
rangeSum += rangeSumBST(root->right, low, high);
return rangeSum;
}
};Python 範例程式碼
# 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:
# Init
ans = root.val if root.val >= low and root.val <= high else 0
# Iter
if root.left and root.val > low:
ans += self.rangeSumBST(root.left, low, high)
if root.right and root.val < high:
ans += self.rangeSumBST(root.right, low, high)
return ans