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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
SOLUTION 1:
使用递归解决,根据left是否为空,先连接left tree, 然后再连接右子树。使用一个tail 来记录链的结尾。在递归之前,先将root.left,root.right保存下来。
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 public void flatten(TreeNode root) { 12 dfs(root); 13 } 14 15 // return : the tail of the list. 16 public TreeNode dfs(TreeNode root) { 17 if (root == null) { 18 return null; 19 } 20 21 TreeNode left = root.left; 22 TreeNode right = root.right; 23 24 // Init the root. 25 root.left = null; 26 root.right = null; 27 28 TreeNode tail = root; 29 30 // connect the left tree. 31 if (left != null) { 32 tail.right = left; 33 tail = dfs(left); 34 } 35 36 // connect the right tree. 37 if (right != null) { 38 tail.right = right; 39 tail = dfs(right); 40 } 41 42 return tail; 43 } 44 }
Leetcode:Flatten Binary Tree to Linked List 解题报告
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原文地址:http://www.cnblogs.com/yuzhangcmu/p/4186572.html