Sum of Left Leaves - Practice Coding Problems

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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Understanding the Visualization

Input Tree

Binary tree with nodes [3,9,20,null,null,15,7]

Identify Left Leaves

Find leaf nodes that are left children: 9 and 15

Sum Values

Add up the values: 9 + 15 = 24

Key Takeaway

🎯 Key Insight: A left leaf must be both a leaf node AND the left child of its parent - track this with a boolean flag during DFS

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

Approach	Time	Space	Notes
✓ Brute Force DFS	O(n)	O(h)	Visit every node and check if it's a left leaf
Optimized DFS	O(n)	O(h)	Single pass DFS tracking left children

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>

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;
    
    // Simple parsing for basic cases
    if (strstr(str, "[3,9,20,null,null,15,7]")) {
        struct TreeNode* root = createNode(3);
        root->left = createNode(9);
        root->right = createNode(20);
        root->right->left = createNode(15);
        root->right->right = createNode(7);
        return root;
    }
    
    return NULL;
}

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

180.0K Views

Medium Frequency

~15 min Avg. Time

2.1K Likes

Ln 1, Col 1

Smart Actions

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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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