[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].

两棵二叉搜索树中的所有元素。题目就是题意，题目给的是两棵二叉搜索树，请你返回的是一个有序的list，里面包含了来自两棵树的所有元素。

这道题的思路就是二叉树的中序遍历94题 + merge two lists 21题。因为input给的是二叉搜索树所以最好是用中序遍历。

时间O(n)

空间O(n)

Java实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {
        List<Integer> list1 = new ArrayList<>();
        inorder(root1, list1);
        List<Integer> list2 = new ArrayList<>();
        inorder(root2, list2);
        return merge(list1, list2);
    }

    private void inorder(TreeNode root, List<Integer> list) {
        if (root == null) {
            return;
        }
        inorder(root.left, list);
        list.add(root.val);
        inorder(root.right, list);
    }

    private List<Integer> merge(List<Integer> list1, List<Integer> list2) {
        List<Integer> res = new ArrayList<>();
        int i = 0;
        int j = 0;
        while (i < list1.size() && j < list2.size()) {
            if (list1.get(i) < list2.get(j)) {
                res.add(list1.get(i));
                i++;
            } else {
                res.add(list2.get(j));
                j++;
            }
        }
        while (i < list1.size()) {
            res.add(list1.get(i));
            i++;
        }
        while (j < list2.size()) {
            res.add(list2.get(j));
            j++;
        }
        return res;
    }
}

 

LeetCode 题目总结