2018-09-01 23:04:30 +0000   |     algorithm leetcode binary search tree tree binary tree   |   Viewed times   |    

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

in-order遍历二叉树

然后根据拿到的数值,重新构造一棵新的二叉树。

代码

class Solution {
    public TreeNode increasingBST(TreeNode root) {
        dummy = new TreeNode(0);
        TreeNode head = dummy;
        inOrder(root);
        return head.right;   
    }
    private TreeNode dummy;
    private void inOrder(TreeNode root) {
        if (root == null) {
            return;
        }
        inOrder(root.left);
        dummy.right = new TreeNode(root.val);
        dummy = dummy.right;
        inOrder(root.right);
    }
}