二叉搜索树中序遍历

难度简单

题目描述

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。

解题思路

参考：剑指 Offer 36. 二叉搜索树与双向链表

Code

/**
 * 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 head, pre;
    public TreeNode increasingBST(TreeNode root) {
        if (root == null) {
            return head;
        }
        recurBST(root);
        return head;
    }
    public void recurBST(TreeNode cur) {
        if (cur == null) {
            return;
        }
        recurBST(cur.left);
        if (pre == null) {
            head = cur;
        } else {
            pre.right = cur;
        }
        cur.left = null;
        pre = cur;
        recurBST(cur.right);
    }
}