题解 | 合并二叉树

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param t1 TreeNode类 
     * @param t2 TreeNode类 
     * @return TreeNode类
     */
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if (!t1)  return t2;
        if (!t2)  return t1;
        t1->val+=t2->val;
        t1->left = mergeTrees(t1->left,t2->left);
        t1->right = mergeTrees(t1->right,t2->right);
        return t1;
    }
};
//时间复杂度o(n)  空间复杂度o(1)

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def mergeTrees(t1: TreeNode, t2: TreeNode) -> TreeNode:
    if not t1:
        return t2
    if not t2:
        return t1
    # 两个节点都存在
    new_node = TreeNode(t1.val + t2.val)
    new_node.left = mergeTrees(t1.left, t2.left)
    new_node.right = mergeTrees(t1.right, t2.right)
    return new_node
//创建新树 ,时间复杂度o(n)  空间复杂度o(n)