leetCode164|递增顺序查找树

示例 ：

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

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

输出：[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  
 

提示：

给定树中的结点数介于 1 和 100 之间。
每个结点都有一个从 0 到 1000 范围内的唯一整数值。

 

import java.util.ArrayList;
import java.util.List;

public  class IncreasingBSTTest {
    public static void main(String[] args) {
        TreeNode t1 = new TreeNode(5);
        TreeNode t2 = new TreeNode(3);
        TreeNode t3 = new TreeNode(6);
        TreeNode t4 = new TreeNode(2);
        TreeNode t5 = new TreeNode(4);
        TreeNode t6 = new TreeNode(8);
        TreeNode t7 = new TreeNode(1);
        TreeNode t8 = new TreeNode(7);
        TreeNode t9 = new TreeNode(9);
        t1.left = t2;
        t1.right = t3;
        t2.left = t4;
        t2.right = t5;
        t3.right = t6;
        t4.left = t7;
        t6.left = t8;
        t6.right = t9;
        increasingBST(t1);


    }

    public static TreeNode increasingBST(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        if (root == null) {
            return null;
        }
        dfs(root, list);
        TreeNode dummyNode = new TreeNode(-1);
        TreeNode tempNode = dummyNode;
        for (Integer integer : list) {
            TreeNode temp = new TreeNode(integer);
            tempNode.right = temp;
            tempNode = tempNode.right;
        }
        return dummyNode.right;
    }

    private static void dfs(TreeNode root, List<Integer> list) {
        if (root == null) {
            return;
        }
        if (root.left != null) {
            dfs(root.left, list);
        }
        list.add(root.val);
        if (root.right != null) {
            dfs(root.right, list);
        }
    }
}