Question

938. Range Sum of BST

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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.

# 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;
  }
}