Sum of Root To Leaf Binary Numbers

Sum of Root To Leaf Binary Numbers - Problem

You are given the root of a binary tree where each node has a value 0 or 1. Each root-to-leaf path represents a binary number starting with the most significant bit.

For example, if the path is 0 → 1 → 1 → 0 → 1, then this could represent 01101 in binary, which is 13 in decimal.

For all leaves in the tree, consider the numbers represented by the path from the root to that leaf. Return the sum of these numbers.

The test cases are generated so that the answer fits in a 32-bits integer.

Input & Output

Example 1 — Basic Tree

$ Input: root = [1,0,1,0,1,0,1]

› Output: 22

💡 Note: Four root-to-leaf paths: (1→0→0)=4, (1→0→1)=5, (1→1→0)=6, (1→1→1)=7. Sum: 4+5+6+7=22

Example 2 — Simple Tree

$ Input: root = [0]

› Output: 0

💡 Note: Single node with value 0, representing binary number 0, so sum is 0

Example 3 — Linear Path

$ Input: root = [1,1,null]

› Output: 3

💡 Note: One path: (1→1) represents binary 11 = decimal 3

Constraints

The number of nodes in the tree is in the range [1, 1000].
Node.val is 0 or 1.

Visualization

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Asked in

f Facebook 25 a Amazon 20 G Google 15 M Microsoft 12

The optimal solution uses DFS to build binary numbers incrementally while traversing the tree. At each node, we calculate current_value = previous_value * 2 + node.val. When reaching leaves, we add the complete value to our sum. Time: O(n), Space: O(h) where h is tree height.

Common Approaches

✓ Brute Force - Collect All Paths

⏱️ Time: O(n) Space: O(h × p)

First traverse the tree to collect all root-to-leaf paths as binary strings. Then convert each binary string to its decimal equivalent and sum them all up. This requires two separate phases.

DFS with Running Sum

⏱️ Time: O(n) Space: O(h)

Use DFS to traverse the tree while building the binary number on-the-fly. At each node, multiply the current value by 2 and add the node's bit. When reaching a leaf, add the complete number to the total sum.

Brute Force - Collect All Paths — Algorithm Steps

Step 1: DFS to find all root-to-leaf paths as binary strings
Step 2: Convert each binary string to decimal
Step 3: Sum all decimal values

Visualization

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Step-by-Step Walkthrough

Find All Paths

DFS traversal collects binary strings: ['101', '100', '110', '111']

Convert Each

Convert each binary string: 101→5, 100→4, 110→6, 111→7

Sum Results

Add all decimal values: 5 + 4 + 6 + 7 = 22

Code -

solution.c — C

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

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

static char paths[1000][1000];
static int pathCount = 0;

void dfs(struct TreeNode* node, char* path) {
    if (!node) return;
    
    int len = strlen(path);
    sprintf(path + len, "%d", node->val);
    
    if (!node->left && !node->right) {
        strcpy(paths[pathCount], path);
        pathCount++;
        path[len] = '\0'; // backtrack
        return;
    }
    
    dfs(node->left, path);
    dfs(node->right, path);
    path[len] = '\0'; // backtrack
}

int binaryToDecimal(const char* binaryStr) {
    int total = 0;
    int len = strlen(binaryStr);
    for (int i = 0; i < len; i++) {
        total = total * 2 + (binaryStr[i] - '0');
    }
    return total;
}

int solution(struct TreeNode* root) {
    if (!root) return 0;
    
    pathCount = 0;
    char path[1000] = "";
    dfs(root, path);
    
    int total = 0;
    for (int i = 0; i < pathCount; i++) {
        total += binaryToDecimal(paths[i]);
    }
    
    return total;
}

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

struct TreeNode* buildTree(char nodes[][10], int size) {
    if (size == 0 || strcmp(nodes[0], "null") == 0) return NULL;
    
    struct TreeNode* root = createNode(atoi(nodes[0]));
    static struct TreeNode* queue[10000];
    int front = 0, rear = 0;
    queue[rear++] = root;
    int i = 1;
    
    while (front < rear && i < size) {
        struct TreeNode* node = queue[front++];
        
        if (i < size && strcmp(nodes[i], "null") != 0) {
            node->left = createNode(atoi(nodes[i]));
            queue[rear++] = node->left;
        }
        i++;
        
        if (i < size && strcmp(nodes[i], "null") != 0) {
            node->right = createNode(atoi(nodes[i]));
            queue[rear++] = node->right;
        }
        i++;
    }
    
    return root;
}

void parseArray(const char* str, char nodes[][10], int* size) {
    *size = 0;
    const char* p = str;
    
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        
        int j = 0;
        if (strncmp(p, "null", 4) == 0) {
            strcpy(nodes[*size], "null");
            p += 4;
        } else {
            while (*p && *p != ',' && *p != ']' && *p != ' ') {
                nodes[*size][j++] = *p++;
            }
            nodes[*size][j] = '\0';
        }
        (*size)++;
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    static char nodes[1000][10];
    int nodeCount;
    parseArray(line, nodes, &nodeCount);
    
    struct TreeNode* root = buildTree(nodes, nodeCount);
    int result = solution(root);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Visit each node once to build paths, then convert each path

✓ Linear Growth

Space Complexity

O(h × p)

Store all paths where h is height and p is number of paths

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Collect All Paths — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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