AlgorithmsEasy DifficultyIterative DFSJavaLeetCode SolutionsRecursive DFSStackTrees

# [LeetCode 404] Sum of Left Leaves

Find the sum of all left leaves in a given binary tree.

Example:

```    3
/ \
9  20
/  \
15   7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.```

## RECURSIVE DFS

```/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
/*
Basic idea: the sum of a node = f(leftSubTree) + f(rightSubtree)
However we need to distinguish between left leaf or right leaf
*/

public class Solution {
public int sumOfLeftLeaves(TreeNode root) {
return helper(root, false);
}

private int helper(TreeNode root, boolean isLeft){
if(root == null) return 0;
/*This node is a leaf, if it's a left leaf, we return the value
if it's a right leaf we return 0 since right leaf doesn't count*/
if(root.left == null && root.right == null){
if(isLeft){
return root.val;
}
return 0;
}
return helper(root.left, true) + helper(root.right, false);
}

}```

## ITERATIVE DFS

```/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public int sumOfLeftLeaves(TreeNode root) {
if(root == null){
return 0;
}

Stack<TreeNode> stack = new Stack<>();

stack.push(root);
int sum =0;

while(!stack.isEmpty()){

TreeNode node = stack.pop();

if(node.left!=null){
if(node.left.left==null&&node.left.right==null){
sum += node.left.val;
}else{
stack.push(node.left);
}
}

if(node.right!=null){
if(node.right.left!=null||node.right.right!=null){
stack.push(node.right);
}
}

}

return sum;
}
}```
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