Convert a Binary Tree into a Sum Tree

Problem and examples

Given a binary tree, convert it into a Sum Tree. In a Sum Tree, each node is assigned a new value which is the sum of the values of its left and right subtrees in the original tree. The value of a leaf node becomes 0 because it has no children. If the tree is empty, return null or handle it as a base case.

Input Tree	Level Order Output	Description
[10, -2, 6, 8, -4, null, 5]	[[13], [4, 5], [0, 0, 0]]	Standard tree: Converted to sum tree where each node is the sum of values in its left and right subtrees
[1]	[[0]]	Edge case: A single node becomes 0 as it has no children
[]	[]	Edge case: Empty tree
[5, 3, null, 2, null, null, null, 1]	[[6], [3], [1], [0]]	Deep left-skewed tree converted into a sum tree; each node is updated recursively
[7, null, 9, null, null, null, 11]	[[20], [11], [0]]	Right-skewed tree: Each node updated to sum of subtree

How it works

Understanding the Problem

The task is to convert a given binary tree into a sum tree. In a sum tree, each node’s value is replaced by the sum of values of its left and right subtrees in the original tree. For leaf nodes, since they don’t have children, their value becomes 0.

This transformation is done recursively, which means for each node, we first compute the sum of its left and right subtrees and update the node’s value, then repeat the process for its children.

Step-by-Step Solution with Example

Step 1: Understand Tree Structure

Consider this example binary tree:


        10
       /        -2    6
     /    /     8  -4 7   5

We need to convert this into a sum tree. For every node, we’ll replace its value with the sum of values in its left and right subtrees from the original tree.

Step 2: Process Leaf Nodes

Leaf nodes have no children, so their new value becomes 0.

8 → 0
-4 → 0
7 → 0
5 → 0

Step 3: Process Intermediate Nodes

Now, we go one level up and calculate the new values based on the original values of children.

-2 has children 8 and -4 → new value = 8 + (-4) = 4
6 has children 7 and 5 → new value = 7 + 5 = 12

Step 4: Process Root Node

Finally, we compute the value for the root node.

10 has children -2 and 6 → in original tree, their values were -2 and 6 → new value = -2 + 6 = 4

Step 5: Update Tree

After the transformation, the sum tree looks like this:


        4
       /        4    12
     /    /     0   0 0   0

Edge Cases

Case 1: Tree with Multiple Levels

This is the general case like the example above. Each internal node gets updated based on the sum of its children’s original values. Leaf nodes become 0.

Case 2: Tree with a Single Node

If the tree only has one node, there are no children to sum. Therefore, the single node becomes 0, since it's considered a leaf node.

Case 3: Empty Tree

If the input tree is empty (i.e., the root is null), then the output is also an empty tree or null. This is the base case in our recursion.

Finally

Converting a binary tree to a sum tree requires a post-order traversal, where we visit left and right children before updating the current node. The solution handles all edge cases and maintains correctness by updating nodes only after their subtrees are processed. Always remember to handle base cases like leaf nodes and empty trees to avoid runtime errors.

If the tree is empty, return 0.
Recursively convert the left subtree into a sum tree and obtain its sum.
Recursively convert the right subtree into a sum tree and obtain its sum.
Store the current node's original value in a temporary variable, e.g., oldVal.
Update the current node's value to the sum of the left and right subtree sums.
Return the sum of the updated node value and the original value (node.val + oldVal) so it can be used by the parent node.

Implementation and efficiency

C implementation

#include <stdio.h>
#include <stdlib.h>

typedef struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
} TreeNode;

TreeNode* createNode(int val) {
    TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode));
    node->val = val;
    node->left = node->right = NULL;
    return node;
}

int toSumTree(TreeNode* root) {
    if (root == NULL) return 0;
    int oldVal = root->val;
    int leftSum = toSumTree(root->left);
    int rightSum = toSumTree(root->right);
    root->val = leftSum + rightSum;
    return root->val + oldVal;
}

int main() {
    TreeNode* root = createNode(10);
    root->left = createNode(-2);
    root->left->left = createNode(8);
    root->left->right = createNode(-4);
    root->right = createNode(6);
    toSumTree(root);
    printf("Root value after conversion: %d\n", root->val);
    return 0;
}

C++ implementation

#include <iostream>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left, *right;
    TreeNode(int x): val(x), left(nullptr), right(nullptr) {}
};

int toSumTree(TreeNode* root) {
    if (root == nullptr) return 0;
    int oldVal = root->val;
    int leftSum = toSumTree(root->left);
    int rightSum = toSumTree(root->right);
    root->val = leftSum + rightSum;
    return root->val + oldVal;
}

int main() {
    TreeNode* root = new TreeNode(10);
    root->left = new TreeNode(-2);
    root->left->left = new TreeNode(8);
    root->left->right = new TreeNode(-4);
    root->right = new TreeNode(6);
    toSumTree(root);
    cout << "Root value after conversion: " << root->val << endl;
    return 0;
}

Python implementation

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def to_sum_tree(root):
    if root is None:
        return 0
    old_val = root.val
    left_sum = to_sum_tree(root.left)
    right_sum = to_sum_tree(root.right)
    root.val = left_sum + right_sum
    return root.val + old_val

if __name__ == '__main__':
    root = TreeNode(10, TreeNode(-2, TreeNode(8), TreeNode(-4)), TreeNode(6))
    to_sum_tree(root)
    print("Root value after conversion:", root.val)

Java implementation

class TreeNode {
    int val;
    TreeNode left, right;
    TreeNode(int val) { this.val = val; }
}

public class SumTreeJava {
    public int toSumTree(TreeNode root) {
        if (root == null) return 0;
        int oldVal = root.val;
        int leftSum = toSumTree(root.left);
        int rightSum = toSumTree(root.right);
        root.val = leftSum + rightSum;
        return root.val + oldVal;
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(10);
        root.left = new TreeNode(-2);
        root.left.left = new TreeNode(8);
        root.left.right = new TreeNode(-4);
        root.right = new TreeNode(6);
        SumTreeJava obj = new SumTreeJava();
        obj.toSumTree(root);
        System.out.println("Root value after conversion: " + root.val);
    }
}

JavaScript implementation

class TreeNode {
    constructor(val, left = null, right = null) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

function toSumTree(root) {
    if (!root) return 0;
    const oldVal = root.val;
    const leftSum = toSumTree(root.left);
    const rightSum = toSumTree(root.right);
    root.val = leftSum + rightSum;
    return root.val + oldVal;
}

const root = new TreeNode(10, new TreeNode(-2, new TreeNode(8), new TreeNode(-4)), new TreeNode(6));
toSumTree(root);
console.log("Root value after conversion:", root.val);

Go implementation

package main

import (
    "fmt"
)

type TreeNode struct {
    val   int
    left  *TreeNode
    right *TreeNode
}

func toSumTree(root *TreeNode) int {
    if root == nil {
        return 0
    }
    oldVal := root.val
    leftSum := toSumTree(root.left)
    rightSum := toSumTree(root.right)
    root.val = leftSum + rightSum
    return root.val + oldVal
}

func main() {
    root := &TreeNode{val: 10,
        left: &TreeNode{val: -2, left: &TreeNode{val: 8}, right: &TreeNode{val: -4}},
        right: &TreeNode{val: 6}}
    toSumTree(root)
    fmt.Println("Root value after conversion:", root.val)
}

Rust implementation

struct TreeNode {
    val: i32,
    left: Option<Box<TreeNode>>,
    right: Option<Box<TreeNode>>,
}

impl TreeNode {
    fn new(val: i32) -> Self {
        TreeNode { val, left: None, right: None }
    }
}

fn to_sum_tree(node: &mut Option<Box<TreeNode>>) -> i32 {
    if let Some(ref mut n) = node {
        let old_val = n.val;
        let left_sum = to_sum_tree(&mut n.left);
        let right_sum = to_sum_tree(&mut n.right);
        n.val = left_sum + right_sum;
        return n.val + old_val;
    }
    0
}

fn main() {
    let mut root = Some(Box::new(TreeNode {
        val: 10,
        left: Some(Box::new(TreeNode {
            val: -2,
            left: Some(Box::new(TreeNode::new(8))),
            right: Some(Box::new(TreeNode::new(-4)))
        })),
        right: Some(Box::new(TreeNode::new(6)))}));

    to_sum_tree(&mut root);
    if let Some(node) = root { println!("Root value after conversion: {}", node.val); }
}

Kotlin implementation

class TreeNode(var val: Int) {
    var left: TreeNode? = null
    var right: TreeNode? = null
}

fun toSumTree(root: TreeNode?): Int {
    if (root == null) return 0
    val oldVal = root.val
    val leftSum = toSumTree(root.left)
    val rightSum = toSumTree(root.right)
    root.val = leftSum + rightSum
    return root.val + oldVal
}

fun main() {
    val root = TreeNode(10)
    root.left = TreeNode(-2)
    root.left?.left = TreeNode(8)
    root.left?.right = TreeNode(-4)
    root.right = TreeNode(6)
    toSumTree(root)
    println("Root value after conversion: ${root.val}")
}

TypeScript implementation

class TreeNode {
    val: number;
    left: TreeNode | null;
    right: TreeNode | null;

    constructor(val: number, left: TreeNode | null = null, right: TreeNode | null = null) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

function toSumTree(root: TreeNode | null): number {
    if (root === null) return 0;
    const oldVal = root.val;
    const leftSum = toSumTree(root.left);
    const rightSum = toSumTree(root.right);
    root.val = leftSum + rightSum;
    return root.val + oldVal;
}

const root = new TreeNode(10, new TreeNode(-2, new TreeNode(8), new TreeNode(-4)), new TreeNode(6));
toSumTree(root);
console.log("Root value after conversion:", root.val);