[LeetCode] 938. Range Sum of BST

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:

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:

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.

二叉搜索树的范围和。

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。

思路

这道题用迭代或者用递归做都可以，我直接给出代码。两种做法时间空间复杂度一样，均为O(n)。其中因为是一棵二叉搜索树的关系，我们可以利用BST的性质进行合理剪枝。

复杂度

时间O(n)
空间O(n)

代码

迭代/BFS代码

/**
 * 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) {
		// corner case
		if (root == null) {
			return 0;
		}

		// normal case
		int sum = 0;
		Queue<TreeNode> queue = new LinkedList<>();
		queue.offer(root);
		while (!queue.isEmpty()) {
			TreeNode cur = queue.poll();
			if (cur.val >= low && cur.val <= high) {
				sum += cur.val;
			}
			if (cur.left != null) {
				queue.offer(cur.left);
			}
			if (cur.right != null) {
				queue.offer(cur.right);
			}
		}
		return sum;
	}
}

递归代码

/**
 * 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) {
        if (root == null) {
            return 0;
        }
        if (root.val > high) {
            return rangeSumBST(root.left, low, high);
        }
        if (root.val < low) {
            return rangeSumBST(root.right, low, high);
        }
        return root.val + rangeSumBST(root.left, low, high) + rangeSumBST(root.right, low, high);
    }
}