[LC] 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.

Solution 1:
BFS

/**
 * 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) {
        int res = 0;
        if (root == null) {
            return res;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            if (cur.left != null) {
                if (cur.left.left == null && cur.left.right == null) {
                    res += cur.left.val;
                } else {
                    queue.add(cur.left);
                }
            }
            if (cur.right != null) {
                queue.add(cur.right);
            }
        }
        return res;
    }
}

Solution 2:
DFS

/**
 * 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) {
            if (root.left.left == null && root.left.right == null) {
                res += root.left.val;
            } else {
                res += sumOfLeftLeaves(root.left);
            }
        }
        res += sumOfLeftLeaves(root.right);
        return res;
    }
}