【算法与数据结构】617、LeetCode合并二叉树

所有的LeetCode题解索引，可以看这篇文章——【算法和数据结构】LeetCode题解。

一、题目

二、解法

  思路分析：采用递归的方式遍历二叉树，【算法与数据结构】144、94、145LeetCode二叉树的前中后遍历（递归法、迭代法），递归法程序可以参考这篇文章。递归重要的是三步骤：输入参数和返回值；终止条件；单层递归逻辑。
  程序如下：

class Solution {
public:
    //1、 输入参数root1 root2
	TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
        // 2、终止条件
        if (!root1) return root2;
        if (!root2) return root1;

        // 3、单层递归逻辑
        root1->val += root2->val;
        root1->left = mergeTrees(root1->left, root2->left);
        root1->right = mergeTrees(root1->right, root2->right);

        // 1、返回值 root1
        return root1;
	}
};
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三、完整代码

# include 
# include 
# include 
# include 
# include 
using namespace std;

// 树节点定义
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:
    //1、 输入参数root1 root2
    TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
        // 2、终止条件
        if (!root1) return root2;
        if (!root2) return root1;

        // 3、单层递归逻辑
        root1->val += root2->val;
        root1->left = mergeTrees(root1->left, root2->left);
        root1->right = mergeTrees(root1->right, root2->right);

        // 1、返回值 root1
        return root1;
    }
};

// 前序遍历迭代法创建二叉树，每次迭代将容器首元素弹出（弹出代码还可以再优化）
void Tree_Generator(vector<string>& t, TreeNode*& node) {
    if (!t.size() || t[0] == "NULL") return;    // 退出条件
    else {
        node = new TreeNode(stoi(t[0].c_str()));    // 中
        if (t.size()) {
            t.assign(t.begin() + 1, t.end());
            Tree_Generator(t, node->left);              // 左
        }
        if (t.size()) {
            t.assign(t.begin() + 1, t.end());
            Tree_Generator(t, node->right);             // 右
        }
    }
}

template<typename T>
void my_print(T& v, const string msg)
{
    cout << msg << endl;
    for (class T::iterator it = v.begin(); it != v.end(); it++) {
        cout << *it << ' ';
    }
    cout << endl;
}

template<class T1, class T2>
void my_print2(T1& v, const string str) {
    cout << str << endl;
    for (class T1::iterator vit = v.begin(); vit < v.end(); ++vit) {
        for (class T2::iterator it = (*vit).begin(); it < (*vit).end(); ++it) {
            cout << *it << ' ';
        }
        cout << endl;
    }
}

// 层序遍历
vector<vector<int>> levelOrder(TreeNode* root) {
    queue<TreeNode*> que;
    if (root != NULL) que.push(root);
    vector<vector<int>> result;
    while (!que.empty()) {
        int size = que.size();  // size必须固定, que.size()是不断变化的
        vector<int> vec;
        for (int i = 0; i < size; ++i) {
            TreeNode* node = que.front();
            que.pop();
            vec.push_back(node->val);
            if (node->left) que.push(node->left);
            if (node->right) que.push(node->right);
        }
        result.push_back(vec);
    }
    return result;
}

int main()
{
    vector<string> t1 = { "1", "3", "5", "NULL", "NULL", "NULL", "2", "NULL", "NULL" };   // 前序遍历
    my_print(t1, "目标树");
    TreeNode* root1 = new TreeNode();
    Tree_Generator(t1, root1);
    vector<vector<int>> tree1 = levelOrder(root1);
    my_print2<vector<vector<int>>, vector<int>>(tree1, "目标树:");

    vector<string> t2 = { "2", "1", "NULL", "4", "NULL", "NULL", "3", "NULL", "7", "NULL", "NULL" };   // 前序遍历
    my_print(t2, "目标树");
    TreeNode* root2 = new TreeNode();
    Tree_Generator(t2, root2);
    vector<vector<int>> tree2 = levelOrder(root2);
    my_print2<vector<vector<int>>, vector<int>>(tree2, "目标树:");

    Solution s;
    TreeNode* root = s.mergeTrees(root1, root2);
    vector<vector<int>> tree = levelOrder(root);
    my_print2<vector<vector<int>>, vector<int>>(tree, "目标树:");

    system("pause");
    return 0;
}
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end