一:题目

二:上码

/*** 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 {/**1.前序遍历2.递归终止条件:(只考虑单层循环)当我们遍历到某个二叉树的某个结点为空时，返回另一个二叉树的在该位置对应的结点3.递归体的话  我们是选择root1上进行填补*///单层递归逻辑public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {if (root1 == null) return root2;if (root2 == null) return root1;root1.val += root2.val;root1.left = mergeTrees(root1.left,root2.left);root1.right = mergeTrees(root1.right,root2.right);    return root1;}
}