问题抽象

合并两颗二叉树，重叠部分值相加

算法

直接实现即可，容易想到以某棵树为准，将另一颗树上的节点合并过去

复杂度

$O(N)$

代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
        root1 = dfs(root1, root2);
        return root1;
    }
    TreeNode* dfs(TreeNode* r1, TreeNode *r2){
        if(!r1 && !r2) return nullptr;
        if(!r1 && r2){
            return r2;
        }

        if(r1 && r2){
            r1 -> val += r2 -> val;
            r1->left = dfs(r1->left, r2->left);
            r1 -> right = dfs(r1->right, r2->right);
        }

        return r1;
    }
};