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938. 二叉搜索树的范围和

English Version

题目描述

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

解法

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: TreeNode, low: int, high: int) -> int:
        def search(node):
            if not node:
                return
            if low <= node.val <= high:
                self.ans += node.val
                search(node.left)
                search(node.right)
            elif node.val < low:
                search(node.right)
            elif node.val > high:
                search(node.left)

        self.ans = 0
        search(root)
        return self.ans

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

C++

/**
 * 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) {
        if (root == nullptr) return 0;
        if (low <= root->val && root->val <= high) {
            return root->val + rangeSumBST(root->left, low, high) + rangeSumBST(root->right, low, high);
        } else if (root->val < low) {
            return rangeSumBST(root->right, low, high);
        } else {
            return rangeSumBST(root->left, low, high);
        }
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func rangeSumBST(root *TreeNode, low int, high int) int {
	if root == nil {
		return 0
	}
	if low <= root.Val && root.Val <= high {
		return root.Val + rangeSumBST(root.Left, low, high) + rangeSumBST(root.Right, low, high)
	} else if root.Val < low {
		return rangeSumBST(root.Right, low, high)
	} else {
		return rangeSumBST(root.Left, low, high)
	}
}

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