题目:
Given a binary tree, flatten it to a linked list in-place.
For example,
Given
1 / 2 5 / 3 4 6
The flattened tree should look like:
1 2 3 4 5 6
题意及分析:给出一棵树,要求给出这棵树的先序遍历组成的链表,但是还是用树表示。首先找到根节点左子节点的最右子节点,然后将根节点的右子树移到该点的右节点上;再将根节点的左子节点移到根节点的右子节点上,并将根节点左子树重置为null;然后将根节点向右子节点移动一位,递归即可。
代码:
/** * 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) { if(root==null) return; if(root.left!=null){ TreeNode cur = root.left; while(cur.right!=null){ cur=cur.right; } cur.right=root.right; root.right=root.left; root.left=null; } flatten(root.right); } }