Sum of Left Leaves Easy 0 attempts
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Sum of Left Leaves

Easy TreeBinary TreeDFS LeetCode

You're given the root of a binary tree. A leaf is any node without children. A node counts as a left leaf only if it's a leaf and it's the left child of its parent. Add up the values of every left leaf in the tree and return the total — 0 if there aren't any.

Right leaves don't count, and neither does a node whose only child is on the right.

Input Format
Implement sumOfLeftLeaves(root). Inputs below show JavaScript function arguments; there is no stdin parsing. Trees use level-order arrays with null for missing children. The editor passes nodes with val, left, and right properties; [] becomes null.
Output Format
Return the result described in the statement.

Sample Input and Output

Example 1 · Run Sample

Input
root = [3,9,20,null,null,15,7]
Output
24

Example 2

Input
root = [1]
Output
0
Constraints
  • 1 <= number of nodes <= 1000
  • -1000 <= Node.val <= 1000
Test Cases (2)
Case 1
Args: [[3,9,20,null,null,15,7]] Expected: 24
Case 2
Args: [[1]] Expected: 0
Topics
TreeBinary TreeDFS

DFS

javascript
function sumOfLeftLeaves(root) {
  if (!root) return 0;
  let sum = 0;
  if (root.left && !root.left.left && !root.left.right) sum += root.left.val;
  else sum += sumOfLeftLeaves(root.left);
  sum += sumOfLeftLeaves(root.right);
  return sum;
}

Time: O(n) | Space: O(h)

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JavaScript