## C# || How To Get Total Sum Of Left Leaves In Binary Tree Using C# The following is a module with functions which demonstrates how to get the total sum of left leaves in a binary tree using C#.

1. Sum Of Left Leaves – Problem Statement

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

A leaf node is a node with no children.

Example 1: ``` Input: root = [3,9,20,null,null,15,7] Output: 24 Explanation: There are two left leaves in the binary tree, with values 9 and 15 respectively. ```

Example 2:

``` Input: root =  Output: 0 ```

2. Sum Of Left Leaves – Solution

The following are two solutions which demonstrates how to get the total sum of left leaves in a binary tree.

Both solutions use Depth First Search to explore items at each level.

In both solutions, we traverse the left and right side of the tree, keeping track of which node is the ‘left’ node.

When a leaf is encountered and its a node on the left side, we increment the final result with the value of the node on the left side.

Recursive

``` 2. Sum Of Left Leaves - Recursive - Solution C# // ============================================================================ // Author: Kenneth Perkins // Date: Nov 3, 2021 // Taken From: http://programmingnotes.org/ // File: Solution.cs // Description: Demonstrates how to get the sum of left leaves // ============================================================================ /** * Definition for a binary tree node. * public class TreeNode { * public int val; * public TreeNode left; * public TreeNode right; * public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) { * this.val = val; * this.left = left; * this.right = right; * } * } */ public class Solution { public int SumOfLeftLeaves(TreeNode root) { return Traverse(root, false); } private int Traverse(TreeNode node, bool isLeft) { if (node == null) { return 0; } // We have reached a leaf node if (node.left == null && node.right == null) { return isLeft ? node.val : 0; } // Keep traversing left and right looking for left leaves return Traverse(node.left, true) + Traverse(node.right, false); } }// http://programmingnotes.org/ 12345678910111213141516171819202122232425262728293031323334353637 // ============================================================================//    Author: Kenneth Perkins//    Date:   Nov 3, 2021//    Taken From: http://programmingnotes.org///    File:  Solution.cs//    Description: Demonstrates how to get the sum of left leaves// ============================================================================/** * Definition for a binary tree node. * public class TreeNode { *     public int val; *     public TreeNode left; *     public TreeNode right; *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) { *         this.val = val; *         this.left = left; *         this.right = right; *     } * } */public class Solution {    public int SumOfLeftLeaves(TreeNode root) {        return Traverse(root, false);    }     private int Traverse(TreeNode node, bool isLeft) {        if (node == null) {            return 0;        }        // We have reached a leaf node        if (node.left == null && node.right == null) {            return isLeft ? node.val : 0;        }        // Keep traversing left and right looking for left leaves        return Traverse(node.left, true) + Traverse(node.right, false);    }}// http://programmingnotes.org/ ```

Iterative

``` 2. Sum Of Left Leaves - Iterative - Solution C# // ============================================================================ // Author: Kenneth Perkins // Date: Nov 3, 2021 // Taken From: http://programmingnotes.org/ // File: Solution.cs // Description: Demonstrates how to get the sum of left leaves // ============================================================================ /** * Definition for a binary tree node. * public class TreeNode { * public int val; * public TreeNode left; * public TreeNode right; * public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) { * this.val = val; * this.left = left; * this.right = right; * } * } */ public class Solution { public int SumOfLeftLeaves(TreeNode root) { if (root == null) { return 0; } // Declare stack var stack = new Stack<KeyValuePair<TreeNode, bool>>(); var result = 0; // Add root to stack stack.Push(new KeyValuePair<TreeNode, bool>(root, false)); // Loop through items on the stack while (stack.Count > 0) { // Get current node and value indicating if its a left node var current = stack.Pop(); var node = current.Key; var isLeft = current.Value; // We have reached a leaf node if (node.left == null && node.right == null) { result += isLeft ? node.val : 0; } else { // Keep traversing left and right looking for left leaves if (node.left != null) { stack.Push(new KeyValuePair<TreeNode, bool>(node.left, true)); } if (node.right != null) { stack.Push(new KeyValuePair<TreeNode, bool>(node.right, false)); } } } return result; } }// http://programmingnotes.org/ 1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556 // ============================================================================//    Author: Kenneth Perkins//    Date:   Nov 3, 2021//    Taken From: http://programmingnotes.org///    File:  Solution.cs//    Description: Demonstrates how to get the sum of left leaves// ============================================================================/** * Definition for a binary tree node. * public class TreeNode { *     public int val; *     public TreeNode left; *     public TreeNode right; *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) { *         this.val = val; *         this.left = left; *         this.right = right; *     } * } */public class Solution {    public int SumOfLeftLeaves(TreeNode root) {        if (root == null) {            return 0;        }        // Declare stack        var stack = new Stack<KeyValuePair<TreeNode, bool>>();        var result = 0;         // Add root to stack        stack.Push(new KeyValuePair<TreeNode, bool>(root, false));         // Loop through items on the stack        while (stack.Count > 0) {            // Get current node and value indicating if its a left node            var current = stack.Pop();            var node = current.Key;            var isLeft = current.Value;             // We have reached a leaf node            if (node.left == null && node.right == null) {                result += isLeft ? node.val : 0;            } else {                // Keep traversing left and right looking for left leaves                if (node.left != null) {                    stack.Push(new KeyValuePair<TreeNode, bool>(node.left, true));                }                if (node.right != null) {                    stack.Push(new KeyValuePair<TreeNode, bool>(node.right, false));                }            }        }         return result;    }}// http://programmingnotes.org/ ```

QUICK NOTES:
The highlighted lines are sections of interest to look out for.

The code is heavily commented, so no further insight is necessary. If you have any questions, feel free to leave a comment below.

Once compiled, you should get this as your output for the example cases:

``` 24 0 ```