897. 递增顺序查找树

1. 问题

Given a 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 1 right child.

给定一个树，按中序遍历重新排列树，使树中最左边的结点现在是树的根，并且每个结点没有左子结点，只有一个右子结点。

2. 解题思路

按照中序遍历，并构建一棵新的树

3. 代码示例

/**
 /**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode increasingBST(TreeNode root) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        orderAndAddList(root, list);
        TreeNode newnode = new TreeNode(0);
        TreeNode cur = newnode;
        for(int i = 0; i < list.size(); i++){
            cur = cur.right = new TreeNode(list.get(i));
        }
        return newnode.right;
    }
    public void orderAndAddList(TreeNode root, ArrayList<Integer> list){
        if(root == null) {
            return;
        }
        orderAndAddList(root.left, list);
        list.add(root.val);
        orderAndAddList(root.right, list);
    }
}