---------------🎈🎈题目链接🎈🎈-------------------

解法1 深度优先 递归 前序

/**
 * 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 TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if(root1==null) return root2;
        if(root2==null) return root1;

        TreeNode root = new TreeNode(root1.val+root2.val);
        
        root.left = mergeTrees(root1.left,root2.left);  
        root.right = mergeTrees(root1.right,root2.right);
        
        return root;
    }
}   
    

解法2 采用队列进行层序遍历 挺巧妙的可以再看

在temp1 的基础上改 改好了返回

class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        Queue<TreeNode> myqueue = new LinkedList<>();
        if(root1 == null) return root2;
        if(root2 == null) return root1;
        myqueue.add(root1);
        myqueue.add(root2);

        while(!myqueue.isEmpty()){
            TreeNode temp1 = myqueue.poll();
            TreeNode temp2 = myqueue.poll();
            temp1.val = temp1.val+temp2.val;

            if(temp1.left!=null && temp2.left!= null){ // 如果两棵树左节点都不为空，加入队列
                myqueue.add(temp1.left);
                myqueue.add(temp2.left);
            }

            if(temp1.right!=null && temp2.right!= null){ // 如果两棵树左节点都不为空，加入队列
                myqueue.add(temp1.right);
                myqueue.add(temp2.right);
            }

            if(temp1.left==null){ // 如果temp1的左节点为空，直接赋值temp2.left
                temp1.left = temp2.left;
            }

            if(temp1.right==null){ // 如果temp1的右节点为空，直接赋值temp2.right
                temp1.right = temp2.right;
            }
        }
        return root1;
    }
}