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
     */

    void visit(TreeNode* node1, TreeNode* node2, TreeNode*& node) {
        if (node1 != nullptr && node2 != nullptr)
            node->val = node1->val + node2->val;
        else
            return;

        if (node1->left != nullptr && node2->left == nullptr) {
            node->left = node1->left;
        } else if (node1->left == nullptr && node2->left != nullptr) {
            node->left = node2->left;
        } else if (node1->left == nullptr && node2->left == nullptr) {
            visit(node1->left, node2->left, node->left);
        } else {
            node->left = new TreeNode(0);
            visit(node1->left, node2->left, node->left);
        }

        if (node1->right != nullptr && node2->right == nullptr) {
            node->right = node1->right;
        } else if (node1->right == nullptr && node2->right != nullptr) {
            node->right = node2->right;
        } else if (node1->right == nullptr && node2->right == nullptr) {
            visit(node1->right, node2->right, node->right);
        } else {
            node->right = new TreeNode(0);
            visit(node1->right, node2->right, node->right);
        }

    }

    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        // write code here
        auto newTree = new TreeNode(0);
        visit(t1, t2, newTree);
        return newTree;
    }
};