Binary Tree Leaf Sum(lintcode 481)

Description

Given a binary tree, calculate the sum of leaves.

Example

Given:

    1
   / \
  2   3
 / 
4

return 7.

Interface

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root the root of the binary tree
     * @return an integer
     */
    public int leafSum(TreeNode root) {
        // Write your code 
    }
}

Idea

root == null is absolutely necessary. When a node has only one child and we send another empty child to traverse(), this condition can help the code return.

Solution

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root the root of the binary tree
     * @return an integer
     */
    private int sum = 0;

    public int leafSum(TreeNode root) {
        // Write your code 
        traverse(root);
        return sum;
    }

    private void traverse(TreeNode root) {
        if (root == null) {
            return;
        }
        if (root.left == null && root.right == null) {
            sum += root.val;
            return;
        }

        traverse(root.left);
        traverse(root.right);
    }
}