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leetcode讲解–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:

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.

题目地址
讲解
这道题讲道理本来应该是很简单的,但我犯了一个很愚蠢的错误,就是直接使用原来的树的结点,这样原来的树的结构也被带过来了,导致形成了死递归。其实我只需要原来的结点的值。temp.right = new TreeNode(root.val); 而不是:temp.right = root;
Java 代码
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) {val = x;}
* }
*/
class Solution {
TreeNode result = new TreeNode(0);
TreeNode temp = result;
public TreeNode increasingBST(TreeNode root) {
if(root==null){
return root;
}
if(root.left!=null){
increasingBST(root.left);
}
temp.right = new TreeNode(root.val);
temp = temp.right;
if(root.right!=null){
increasingBST(root.right);
}
return result.right;
}
}

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