左叶子之和

题目：

给定二叉树的根节点 root ，返回所有左叶子之和。

 

 

示例 1：

 

输入: root = [3,9,20,null,null,15,7]
输出: 24
解释: 在这个二叉树中，有两个左叶子，分别是 9 和 15，所以返回 24
示例 2:

输入: root = [1]
输出: 0
 

提示:

节点数在 [1, 1000] 范围内
-1000 <= Node.val <= 1000

public class Solution {

    //[3,9,20,null,null,15,7]
    public static void main(String[] args) {
        Solution solution = new Solution();
        TreeNode head3 = new TreeNode(-3, null, new TreeNode(4, new TreeNode(-6), null));
        TreeNode head4 = new TreeNode(2, new TreeNode(4, new TreeNode(-5), null), new TreeNode(0));
        TreeNode head6 = new TreeNode(15);
        TreeNode head7 = new TreeNode(7);
        TreeNode head2 = new TreeNode(9, null, null);
        TreeNode head5 = new TreeNode(20, head6, head7);

        TreeNode head1 = new TreeNode(3, new TreeNode(9), new TreeNode(20,new TreeNode(15),new TreeNode(7)));
        //[0,2,4,1,null,3,-1,5,1,null,6,null,8]

        System.out.println(solution.sumOfLeftLeaves(head1));
    }

    public int sumOfLeftLeaves(TreeNode root) {
        int i = 0;//统计值。
        if (root.left != null || root.right != null) {
            i = getLeaf(null, root, i);
        }
        return i;
    }

    public int getLeaf(TreeNode lastNode, TreeNode current, int i) {
        if (current.left != null) {//有左节点
            i = getLeaf(current, current.left, i); //继续迭代左节点并且要获取最新的i值
        } else {//没有左和右节点并且是上级的左节点此时才需要增加 不增加继续往下走判断是不是有右节点。
            if ( current.right == null && lastNode.left == current ) {//没有右节点并且是上级的左节点
                return i + current.val;
            }
        }
        //当前节点左节点解析完成开始解析右节点
        if (current.right != null) {//继续循环右节点并且要获取最新的i值
            i = getLeaf(current, current.right, i);
        }
        //没有右节点返回最新i值
        return i;
    }
}

 

PS：本题转自力扣