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.
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]
Example 2:
Input: root = [5,1,7]
Output: [1,null,5,null,7]
Constraints:
- The number of nodes in the given tree will be in the range [1, 100].
- 0 <= Node.val <= 1000
Solution
/**
* 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);
}
}