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.
代码
其实只需要知道二叉搜索树的根节点，左分支都小于其，右分支都大于其，按照这个规律，直接进行中序遍历，获得到数据，重新构造树即可
/**
* 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) {
ArrayList<Integer> list = new ArrayList<Integer>();
fun(root,list);
TreeNode newnode = new TreeNode(0);
TreeNode cur = newnode;
int len = list.size();
for(int i=0;i<len;i++){
cur = cur.right = new TreeNode(list.get(i));
}
return newnode.right;
}
public void fun(TreeNode root,ArrayList<Integer> list){
if(root==null) return;
fun(root.left,list);
list.add(root.val);
fun(root.right,list);
}
}
/**
* 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) {
ArrayList<Integer> list = new ArrayList<Integer>();
fun(root,list);
TreeNode newnode = new TreeNode(0);
TreeNode cur = newnode;
int len = list.size();
for(int i=0;i<len;i++){
cur = cur.right = new TreeNode(list.get(i));
}
return newnode.right;
}
public void fun(TreeNode root,ArrayList<Integer> list){
if(root==null) return;
fun(root.left,list);
list.add(root.val);
fun(root.right,list);
}
}
