//当前节点的左子树不为空 且是叶子节点
root.left != null 
&&root.left.left==null && root.left.right==null

 
/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public int sumOfLeftLeaves(TreeNode root) {if(root == null) {return 0;}int leftSum = sumOfLeftLeaves(root.left);//当前节点的左子树不为空，且为叶子节点if(root.left != null && root.left.left==null && root.left.right==null){//加左叶子的值leftSum += root.left.val;}//右边的分支也传入，去加所在的左叶子节点int rightSum = sumOfLeftLeaves(root.right);//最后左右分支相加int ans = leftSum + rightSum;return ans;}
}```