给你一棵二叉搜索树，请你按中序遍历将其重新排列为一棵递增顺序搜索树，使树中最左边的节点成为树的根节点，并且每个节点没有左子节点，只有一个右子节点。
示例 1：
输入：root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
输出：[1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
示例 2：
输入：root = [5,1,7]
输出：[1,null,5,null,7]
提示：
- 树中节点数的取值范围是 [1, 100]
- 0 <= Node.val <= 1000
解答
/**
* 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 TreeNode increasingBST(TreeNode root) {
TreeNode newRoot = root;
while (newRoot.left != null) newRoot = newRoot.left;
makeInorderTree(root);
return newRoot;
}
private void makeInorderTree(TreeNode node) {
if (node == null) return;
makeInorderTree(node.left);
if (node.left != null) {
TreeNode preNode = node.left;
while (preNode.right != null) preNode = preNode.right;
preNode.right = node;
preNode.left = null;
node.left = null;
}
TreeNode rightTree = node.right;
if (node.right != null) {
TreeNode postNode = node.right;
while (postNode.left != null) postNode = postNode.left;
node.right = postNode;
}
makeInorderTree(rightTree);
}
}
