##### Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/404.html

# 404. Sum of Left Leaves (Easy)

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.


Related Topics:
Tree

## Solution 1. DFS

// OJ: https://leetcode.com/problems/sum-of-left-leaves
// Time: O(N)
// Space: O(H)
class Solution {
private:
int dfs(TreeNode *root, bool isLeftChild) {
if (!root) return 0;
if (!root->left && !root->right && isLeftChild) return root->val;
return dfs(root->left, true) + dfs(root->right, false);
}
public:
int sumOfLeftLeaves(TreeNode* root) {
return dfs(root, false);
}
};

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int sumOfLeftLeaves(TreeNode root) {
if (root == null || root.left == null && root.right == null)
return 0;
int sum = 0;
queue.offer(root);
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
TreeNode left = node.left, right = node.right;
if (isLeaf(left))
sum += left.val;
if (left != null)
queue.offer(left);
if (right != null)
queue.offer(right);
}
return sum;
}

public boolean isLeaf(TreeNode node) {
if (node == null)
return false;
if (node.left == null && node.right == null)
return true;
return false;
}
}

############

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int sumOfLeftLeaves(TreeNode root) {
if (root == null) {
return 0;
}
int res = 0;
if (root.left != null && root.left.left == null && root.left.right == null) {
res += root.left.val;
}
res += sumOfLeftLeaves(root.left);
res += sumOfLeftLeaves(root.right);
return res;
}
}

• // OJ: https://leetcode.com/problems/sum-of-left-leaves
// Time: O(N)
// Space: O(H)
class Solution {
public:
int sumOfLeftLeaves(TreeNode* root, bool isLeft = false) {
if (!root) return 0;
if (!root->left && !root->right) return isLeft ? root->val : 0;
return sumOfLeftLeaves(root->left, true) + sumOfLeftLeaves(root->right);
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
def sumOfLeftLeaves(self, root: TreeNode) -> int:
if root is None:
return 0
res = 0
if root.left and root.left.left is None and root.left.right is None:
res += root.left.val
res += self.sumOfLeftLeaves(root.left)
res += self.sumOfLeftLeaves(root.right)
return res

############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
def sumOfLeftLeaves(self, root):
"""
:type root: TreeNode
:rtype: int
"""

def helper(root, isLeft):
if not root:
return None
left = helper(root.left, True)
right = helper(root.right, False)
ret = 0
if left is None and right is None and isLeft:
return root.val
if left:
ret += left
if right:
ret += right
return ret

ret = helper(root, False)
if ret:
return ret
return 0


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func sumOfLeftLeaves(root *TreeNode) int {
if root == nil {
return 0
}
res := 0
if root.Left != nil && root.Left.Left == nil && root.Left.Right == nil {
res += root.Left.Val
}
res += sumOfLeftLeaves(root.Left)
res += sumOfLeftLeaves(root.Right)
return res
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

const dfs = (root: TreeNode | null, isLeft: boolean) => {
if (!root) {
return 0;
}
const { val, left, right } = root;
if (!left && !right) {
if (isLeft) {
return val;
}
return 0;
}
return dfs(left, true) + dfs(right, false);
};

function sumOfLeftLeaves(root: TreeNode | null): number {
return dfs(root, false);
}


• // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, is_left: bool) -> i32 {
if root.is_none() {
return 0;
}
let node = root.as_ref().unwrap().borrow();
let left = &node.left;
let right = &node.right;
if left.is_none() && right.is_none() {
if is_left {
return node.val;
}
return 0;
}
Self::dfs(left, true) + Self::dfs(right, false)
}

pub fn sum_of_left_leaves(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
Self::dfs(&root, false)
}
}