Given two binary search treesroot1androot2.

Return a list containingall the integersfromboth treessorted inascendingorder.

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 most5000nodes.
• Each node's value is between[-10^5, 10^5].

兩棵二叉搜尋樹中的所有元素。題目就是題意，題目給的是兩棵二叉搜尋樹，請你返回的是一個有序的list，裡面包含了來自兩棵樹的所有元素。

這道題的思路就是二叉樹的中序遍歷

時間O(n)

空間O(n)

Java實現

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
 
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {
18         List<Integer> list1 = new ArrayList<>();
19         inorder(root1, list1);
20         List<Integer> list2 = new ArrayList<>();
21         inorder(root2, list2);
22         return merge(list1, list2);
23     }
24 
25     private void inorder(TreeNode root, List<Integer> list) {
26         if (root == null) {
27             return;
28         }
29         inorder(root.left, list);
30         list.add(root.val);
31         inorder(root.right, list);
32     }
33 
34     private List<Integer> merge(List<Integer> list1, List<Integer> list2) {
35         List<Integer> res = new ArrayList<>();
36         int i = 0;
37         int j = 0;
38         while (i < list1.size() && j < list2.size()) {
39             if (list1.get(i) < list2.get(j)) {
40                 res.add(list1.get(i));
41                 i++;
42             } else {
43                 res.add(list2.get(j));
44                 j++;
45             }
46         }
47         while (i < list1.size()) {
48             res.add(list1.get(i));
49             i++;
50         }
51         while (j < list2.size()) {
52             res.add(list2.get(j));
53             j++;
54         }
55         return res;
56     }
57 }

LeetCode 題目總結