897. Increasing Order Search Tree

Given the root of a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only one right child.

1
2
3
4
Example 1:

Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

ex1

1
2
3
4
Example 2:

Input: root = [5,1,7]
Output: [1,null,5,null,7]

ex2

Constraints:

The number of nodes in the given tree will be in the range [1, 100].
0 <= Node.val <= 1000

Solution

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
/**
 * 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 {
    TreeNode res = null;
    TreeNode next = null;
    public TreeNode increasingBST(TreeNode root) {
        if (root == null) return null;
        dfs(root);
        return res;
    }
    
    void dfs(TreeNode node) {
        if (node == null) return;
        dfs(node.left);
        if (res == null) {
            res = new TreeNode(node.val);
            next = res;
        } else {
            next.right = new TreeNode(node.val);
            next = next.right;
        }
        
        dfs(node.right);
    }
}

Solution#

Solution