LeetCode404. 左叶子之和

☆方法1：递归

方法2：BFS

class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        /**
         * 方法1：递归
         */
        if (root == null) return 0;
        // 对当前节点进行处理
        int count = 0;
        if (root.left != null && root.left.left == null && root.left.right == null) {
            count = root.left.val;
        }
        // 当前节点 + 其左右子树的
        return count + sumOfLeftLeaves(root.left) + sumOfLeftLeaves(root.right);
        /**
         * 方法2：BFS
         */
        /*if (root == null) return 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int count = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();

                if (cur.left != null) {
                    queue.offer(cur.left);
                    // 保证是叶子节点
                    if (cur.left.left == null && cur.left.right == null) {
                        count += cur.left.val;
                    }
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
        }
        return count;
        */
    }
}