LeetCode-1305. All Elements in Two Binary Search Trees

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]​​.

题解：

/**
 * 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 {
public:
    void getNum(TreeNode * &root, vector<int> &v) {
        if (root != NULL) {
            getNum(root->left, v);
            v.push_back(root->val);
            getNum(root->right, v);
        }
    }
    vector<int> getAllElements(TreeNode* root1, TreeNode* root2) {
        vector<int> a, b, res;
        getNum(root1, a);
        getNum(root2, b);
        int l1 = a.size(), l2 = b.size();
        int k1 = 0, k2 = 0;
        while (k1 != l1 && k2 != l2) {
            if (a[k1] < b[k2]) {
                res.push_back(a[k1++]);
            }
            else {
                res.push_back(b[k2++]);
            }
        }
        if (k1 == l1) {
            while (k2 != l2) {
                res.push_back(b[k2++]);
            }
        }
        if (k2 == l2) {
            while (k1 != l1) {
                res.push_back(a[k1++]);
            }
        }
        return res;
    }
};