Level:

  Medium

题目描述:

Given a binary tree, flatten it to a linked list in-place.

For example, given the following tree:

    1
/ \
2 5
/ \ \
3 4 6

The flattened tree should look like:

1
\
2
\
3
\
4
\
5
\
6

思路分析:

  思路一:将二叉树转换为链表,我们可以先序遍历二叉树将节点保存起来,然后依次访问节点构造链表。

  思路二:不使用额外的空间,在遍历的过程中构造链表。

代码:

代码一

/**public class TreeNode{
int val;
TreeNode left;
TreeNode right;
public TreeNode(int x){
val=x;
}
}*/
public class Solution{
List<TreeNode>list=new ArrayList<>();
public void flatten(TreeNode root){
if(root==null)
return;
pre(root);
for(int i=0;i<list.size()-1;i++){
list.get(i).left=null;
list.get(i).right=list.get(i+1);
}
}
public void pre(TreeNode root){
if(root!=null){
list.add(root);
pre(root.left);
pre(root.right);
}
}
}

代码二

public class Solution{
TreeNode pre=null;
public void flatten(TreeNode root){
if(root==null)
return;
flatten(root.right);
flatten(root.left);
root.right=pre;
root.left=null;
pre=root;
}
}

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