题目:Flatten Binary Tree to Linked List
难度:medium
问题描述:
Given a binary tree, flatten it to a linked list in-place.
For example,
Given
1 / \ 2 5 / \ \ 3 4 6The flattened tree should look like:
1 \ 2 \ 3 \ 4 \ 5 \ 6
解题思路:
一种方法是前序遍历,生成新的链表树输出即可。由于要创建大量节点,费时费力,效率低下。
这里使用递归法解题。root的左子树=null,右子树=左子树的递归+右子树的递归。
代码如下:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public void flatten(TreeNode root) {
treetolist(root);
}
private TreeNode treetolist(TreeNode root){
if(root==null) return null;
TreeNode right=treetolist(root.right);
root.right=treetolist(root.left);
root.left=null;
TreeNode temp=root;
while(temp.right!=null){
temp=temp.right;
}
temp.right=right;
return root;
}
}