1305. All Elements in Two Binary Search Trees**

1305. All Elements in Two Binary Search Trees**

https://leetcode.com/problems/all-elements-in-two-binary-search-trees/

题目描述

Given two binary search trees root1 and root2.

Return a list containing all the integers from both trees sorted in ascending order.

Example 1:

Input: root1 = [2,1,4], root2 = [1,0,3]
Output: [0,1,1,2,3,4]

Example 2:

Input: root1 = [0,-10,10], root2 = [5,1,7,0,2]
Output: [-10,0,0,1,2,5,7,10]

Example 3:

Input: root1 = [], root2 = [5,1,7,0,2]
Output: [0,1,2,5,7]

Example 4:

Input: root1 = [0,-10,10], root2 = []
Output: [-10,0,10]

Example 5:

Input: root1 = [1,null,8], root2 = [8,1]
Output: [1,1,8,8]

Constraints:

• Each tree has at most 5000 nodes.
• Each node’s value is between $[-10^5, 10^5]$.

C++ 实现 1

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对 BST 进行中序遍历, 得到的数组是有序的, 最后将两个有序数组进行合并. C++ 实现 1 给出用自己写的合并方法. C++ 实现 2 用标准库提供的 std::merge 就能更为简洁的实现.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    void inorder(TreeNode *root, vector<int> &res) {
        if (!root) return;
        inorder(root->left, res);
        res.push_back(root->val);
        inorder(root->right, res);
    }
public:
    vector<int> getAllElements(TreeNode* root1, TreeNode* root2) {
        vector<int> v1, v2, res;
        inorder(root1, v1);
        inorder(root2, v2);
        int i = 0, j = 0;
        while (i < v1.size() && j < v2.size()) {
            if (v1[i] < v2[j]) res.push_back(v1[i++]);
            else res.push_back(v2[j++]);
        }
        if (i < v1.size()) std::copy(v1.begin() + i, v1.end(), back_inserter(res));
        if (j < v2.size()) std::copy(v2.begin() + j, v2.end(), back_inserter(res));
        return res;
    }
};

C++ 实现 2

使用标准库提供的 std::merge() 方法, 将两个有序数组合并为一个有序数组.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    void inorder(TreeNode *root, vector<int> &res) {
        if (!root) return;
        inorder(root->left, res);
        res.push_back(root->val);
        inorder(root->right, res);
    }
public:
    vector<int> getAllElements(TreeNode* root1, TreeNode* root2) {
        vector<int> v1, v2;
        inorder(root1, v1);
        inorder(root2, v2);
        vector<int> res(v1.size() + v2.size());
        std::merge(v1.begin(), v1.end(), v2.begin(), v2.end(), res.begin());
        return res;
    }
};

来源：CSDN

作者：珍妮的选择

链接：https://blog.csdn.net/Eric_1993/article/details/104563877