938. Range Sum of BST

Description

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.

Solutions

Solution 1: DFS

We design a function $dfs(root)$, which represents the sum of the values of all nodes in the subtree with $root$ as the root, and the values are within the range $[low, high]$. The answer is $dfs(root)$.

The execution logic of the function $dfs(root)$ is as follows:

If $root$ is null, return $0$.
If the value $x$ of $root$ is within the range $[low, high]$, then the initial answer of the function $dfs(root)$ is $x$, otherwise it is $0$.
If $x > low$, it means that there may be nodes in the left subtree of $root$ with values within the range $[low, high]$, so we need to recursively call $dfs(root.left)$ and add the result to the answer.
If $x < high$, it means that there may be nodes in the right subtree of $root$ with values within the range $[low, high]$, so we need to recursively call $dfs(root.right)$ and add the result to the answer.
Finally, return the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the number of nodes in the binary search tree.

Python3JavaC++GoTypeScriptC#

# 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: Optional[TreeNode]) -> int:
            if root is None:
                return 0
            x = root.val
            ans = x if low <= x <= high else 0
            if x > low:
                ans += dfs(root.left)
            if x < high:
                ans += dfs(root.right)
            return ans

        return dfs(root)

/**
 * 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);
    }

    private 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;
    }
}

/**
 * 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) {
        function<int(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return 0;
            }
            int x = root->val;
            int 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);
    }
};

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

/**
 * 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: TreeNode | null): number => {
        if (!root) {
            return 0;
        }
        const { val, left, right } = root;
        let ans = low <= val && val <= high ? val : 0;
        if (val > low) {
            ans += dfs(left);
        }
        if (val < high) {
            ans += dfs(right);
        }
        return ans;
    };
    return dfs(root);
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public int RangeSumBST(TreeNode root, int low, int high) {
        return dfs(root, low, high);
    }

    private 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;
    }
}

938. Range Sum of BST

Description

Solutions

Solution 1: DFS

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