617. 合并二叉树 - 力扣(LeetCode) (leetcode-cn.com)
class Solution {
public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
//// 如果 r1和r2中,只要有一个是null,函数就直接返回
if(root1==null||root2==null)
return root1==null?root2:root1;
//让r1的值 等于 r1和r2的值累加,再递归的计算两颗树的左节点、右节点
root1.val+=root2.val;
root1.left = mergeTrees(root1.left,root2.left);
root1.right = mergeTrees(root1.right,root2.right);
return root1;
}
}