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.

This problem was taken from https://leetcode.com/problems/range-sum-of-bst/

 

 

DFS Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} low
 * @param {number} high
 * @return {number}
 */
var rangeSumBST = function(root, low, high) {
    if(!root) {
        return;
    }
    console.log(root.val);
    var queue = [root];
    var visited = {};
    var result = 0.0;    

    function dfsTraverse(root, low, high) {
    
        var childNodes = [root?.left, root?.right];

        for(const childNode of childNodes) {
          if(childNode) {
            var val = childNode.val;
            console.log(val);
            dfsTraverse(childNode, low, high);            
          }
        }

    }

    dfsTraverse(root, low, high);
    return result;
};


 function TreeNode(val, left, right) {
     this.val = (val===undefined ? 0 : val)
     this.left = (left===undefined ? null : left)
     this.right = (right===undefined ? null : right)
 }

const treeNode = new TreeNode(
    10,
    new TreeNode(
        5,
        new TreeNode(3),
        new TreeNode(7)
    ),
    new TreeNode(
        15,
        null,
        new TreeNode(18)
    )
)


rangeSumBST(treeNode, 7, 15);

 

BFS solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} low
 * @param {number} high
 * @return {number}
 */
var rangeSumBST = function(root, low, high) {
    if(!root) {
        return 0;
    }
    var queue = [root];
    var visited = {};
    var result = 0.0;    
    if(root.val >= low && root.val <= high) {
        result = root.val;
    }
    
    var leftPointer = 0;
    while(leftPointer <= queue.length) {
        var curentNode = queue[leftPointer];
        leftPointer ++;
        var childNodes = [curentNode?.left, curentNode?.right];
        
        for(const childNode of childNodes) {
            if(childNode) {
                console.log(childNode.val)
                if(childNode.val >= low && childNode.val <= high) {
                    result += childNode.val;
                }
                queue.push(childNode);
            }
        }
        
           
    }
    return result;
    
};