897.递增顺序搜索树(简单)

题目链接

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);
    }
}