Problem

Given a binary tree, find the sum of all the nodes which are left as well as leaves nodes.

Examples

Example 1:

graph TD
	3 --- 9:::GreenNode & 20
	20 --- A[15]:::GreenNode & B[7]

classDef GreenNode fill:#0000ff
  
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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:

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Input: root = [1]
Output: 0

Solution

Method 1 - Inorder Traversal

Approach: Approach is quite simple:

  • Do the inorder traversal
  • check if node if the left child and leaf node.
    • If yes then add it to the sum.

See the code for more understanding.

Code

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public static void leftSum(Node root, Node parent) {
	if (root == null) {
		return 0;
	}
	// if left node is present and is a leaf
	if (root.left != null && root.left.left == null && root.left.right == null) {
		return root.left.val + sumOfLeftLeaves(root.right);
	}

	return sumOfLeftLeaves(root.left) + sumOfLeftLeaves(root.right);
}