LeetCode-Increasing Order Search Tree

Description：
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.

Example 1:

Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]

       5
      / \
    3    6
   / \    \
  2   4    8
 /        / \ 
1        7   9

Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

 1
  \
   2
    \
     3
      \
       4
        \
         5
          \
           6
            \
             7
              \
               8
                \
                 9  

Note:

• The number of nodes in the given tree will be between 1 and 100.
• Each node will have a unique integer value from 0 to 1000.

题意：将一颗二叉搜索树转化为一颗每个节点仅有右节点，没有左节点的二叉树，并且节点是有序的；

解法：二叉搜索树是满足根节点的值大于左节点而小于右节点，因此，对二叉搜索树进行中序遍历得到的序列就是用于构造新二叉树的序列；

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    //保存中序遍历的节点值
    private List<Integer> nodes = new ArrayList<>();
    
    public TreeNode increasingBST(TreeNode root) {
        if (root == null) return null;
        getNodes(root);
        TreeNode tree = new TreeNode(nodes.get(0));
        TreeNode treeNext = tree;
        for (int i = 1; i < nodes.size(); i++) {
            TreeNode temp = new TreeNode(nodes.get(i));
            treeNext.right = temp;
            treeNext = temp;
        }
        return tree;
    }
    
    //得到二叉搜索树的中序遍历节点值
    private void getNodes(TreeNode root) {
        if (root.left == null && root.right == null) {
            nodes.add(root.val);
            return;
        }
        if (root.left != null) getNodes(root.left);
        nodes.add(root.val);
        if (root.right != null) getNodes(root.right);
    }
}