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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.
题解：可以看出flatten以后得到的树其实就是原来树的先序遍历，并且所得到的树的左子都有null，右子都是原树先序遍历时的下一个节点。
利用递归先序遍历树即可，用lastNode保存上一个节点的信息，并且在递归过程中要注意保存根节点的右子，因为在递归遍历左子树的过程中，会把根节点的右子指针重新赋值，右子树会丢失。
代码如下：
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 private TreeNode lastNode = null; 12 public void flatten(TreeNode root) { 13 if(root == null) 14 return; 15 16 if(lastNode != null){ 17 lastNode.left = null; 18 lastNode.right = root; 19 } 20 21 lastNode = root; 22 23 TreeNode right = root.right; 24 flatten(root.left); 25 flatten(right); 26 } 27 }
【leetcode刷题笔记】Flatten Binary Tree to Linked List,布布扣,bubuko.com
【leetcode刷题笔记】Flatten Binary Tree to Linked List
标签：style blog color strong io for
原文地址：http://www.cnblogs.com/sunshineatnoon/p/3853596.html
