标签：问题 时间 nts like rsa bin init 边界条件 problems
1、问题描述
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
Hints:
If you notice carefully in the flattened tree, each node‘s right child points to the next node of a pre-order traversal.
2、边界条件：无
3、思路：从示例可以看出，变成list之后是原来的先序遍历结果。那么就采用先序遍历把treenode转为list，再转为树。
4、代码实现
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public void flatten(TreeNode root) { List<TreeNode> result = new ArrayList<>(); preOrder(result, root); TreeNode dummy = new TreeNode(0); dummy.right = root; TreeNode newTree = dummy; for (int i = 0; i < result.size(); i++) { newTree.right = result.get(i); newTree.left = null; newTree = newTree.right; } root = dummy.right; } public void preOrder(List<TreeNode> result, TreeNode root) { if (root == null) { return; } result.add(root); preOrder(result, root.left); preOrder(result, root.right); } }
方法二
My short post order traversal Java solution for share private TreeNode prev = null; public void flatten(TreeNode root) { if (root == null) return; flatten(root.right);//先把右子树flatten，prev=root.right flatten(root.left);//再把左子树flatten root.right = prev;//左子树的最右边是首先flatten的，所以就挂接了prev=root.right root.left = null; prev = root; }
5、时间复杂度：O(N),空间复杂度：O(N)
6、api
leetcode--114. Flatten Binary Tree to Linked List
标签：问题 时间 nts like rsa bin init 边界条件 problems
原文地址：http://www.cnblogs.com/shihuvini/p/7465580.html
