Sum of Left Leaves

Sum of Left Leaves - Problem

Given the root of a binary tree, return the sum of all left leaves.

A leaf is a node with no children. A left leaf is a leaf that is the left child of another node.

Input & Output

Example 1 — Standard Tree

$ Input: root = [3,9,20,null,null,15,7]

› Output: 24

💡 Note: Left leaves are 9 (left child of 3) and 15 (left child of 20). Sum = 9 + 15 = 24

Example 2 — Single Node

$ Input: root = [1]

› Output: 0

💡 Note: Root node has no parent, so it's not a left leaf. No left leaves exist.

Example 3 — Only Right Children

$ Input: root = [1,null,2,null,3]

› Output: 0

💡 Note: All children are right children (2 is right child of 1, 3 is right child of 2). No left leaves.

Constraints

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

Visualization

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Asked in

F Facebook 15 a Amazon 12 G Google 8

The key insight is to use DFS with a boolean flag to track left children. During traversal, when we find a leaf node that is also a left child, we add it to our sum. Optimized DFS solves this in one pass. Time: O(n), Space: O(h).

Common Approaches

✓ Brute Force DFS

⏱️ Time: O(n) Space: O(h)

Traverse the entire tree and for each node, check if it's a leaf and if it's the left child of its parent. Sum all qualifying nodes.

Optimized DFS

⏱️ Time: O(n) Space: O(h)

Use depth-first search with a boolean parameter to track whether current node is a left child. When we find a leaf that is also a left child, add it to the sum.

Brute Force DFS — Algorithm Steps

Step 1: Traverse all nodes using DFS
Step 2: For each node, check if it's a leaf and left child
Step 3: Add qualifying nodes to sum

Visualization

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Step-by-Step Walkthrough

Start DFS

Begin traversal from root, marking left/right children

Check Leaves

For each leaf node, check if it's a left child

Sum Results

Add up all qualifying left leaf values

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;
};

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;
}

int dfs(struct TreeNode* node, int isLeft) {
    if (!node) return 0;
    
    // If it's a leaf and it's a left child
    if (!node->left && !node->right && isLeft) {
        return node->val;
    }
    
    // Continue traversing
    return dfs(node->left, 1) + dfs(node->right, 0);
}

int solution(struct TreeNode* root) {
    if (!root) return 0;
    return dfs(root, 0);
}

struct TreeNode* buildTree(char* str) {
    if (!str || strlen(str) < 3) return NULL;
    
    // Parse the array string
    char* token;
    char* tokens[1000];
    int count = 0;
    
    // Remove brackets and split by comma
    char* copy = malloc(strlen(str) + 1);
    strcpy(copy, str);
    
    // Remove [ and ]
    if (copy[0] == '[') {
        memmove(copy, copy + 1, strlen(copy));
    }
    if (copy[strlen(copy) - 1] == ']') {
        copy[strlen(copy) - 1] = '\0';
    }
    
    token = strtok(copy, ",");
    while (token && count < 1000) {
        // Trim spaces
        while (isspace(*token)) token++;
        char* end = token + strlen(token) - 1;
        while (end > token && isspace(*end)) *end-- = '\0';
        
        tokens[count++] = token;
        token = strtok(NULL, ",");
    }
    
    if (count == 0 || strcmp(tokens[0], "null") == 0) {
        free(copy);
        return NULL;
    }
    
    struct TreeNode* root = createNode(atoi(tokens[0]));
    struct TreeNode* queue[1000];
    int front = 0, rear = 1;
    queue[0] = root;
    
    int i = 1;
    while (front < rear && i < count) {
        struct TreeNode* node = queue[front++];
        
        if (i < count && strcmp(tokens[i], "null") != 0) {
            node->left = createNode(atoi(tokens[i]));
            queue[rear++] = node->left;
        }
        i++;
        
        if (i < count && strcmp(tokens[i], "null") != 0) {
            node->right = createNode(atoi(tokens[i]));
            queue[rear++] = node->right;
        }
        i++;
    }
    
    free(copy);
    return root;
}

void parseArray(const char* str, int* arr, 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;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char input[1000];
    fgets(input, sizeof(input), stdin);
    input[strcspn(input, "\n")] = 0;
    
    struct TreeNode* root = buildTree(input);
    int result = solution(root);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Visit each node once in the tree

✓ Linear Growth

Space Complexity

O(h)

Recursion stack depth equals tree height

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force DFS — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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