[leetcode] 897. Increasing Order Search Tree

题目：

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:

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

代码：

/**
 * 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) {
        List<Integer> order = new ArrayList<>();
        Stack<TreeNode> s = new Stack<>();
        while(root != null || !s.isEmpty()){
            while(root != null){
                s.push(root);
                root = root.left;
            }
            TreeNode node = s.pop();
            order.add(node.val);
            if(node.right != null){
                root = node.right;
            }
        }
        
        TreeNode newRoot = null;
        TreeNode p = null;
        Iterator<Integer> iter = order.iterator();
        while(iter.hasNext()){
            if(newRoot == null){
                newRoot = new TreeNode(iter.next());
                p = newRoot;
            }
            else{
                p.right = new TreeNode(iter.next());
                p = p.right;
            }
        }
        return newRoot;
    }
}