【leetcode刷题笔记】Flatten Binary Tree to Linked List

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：

 /**
  * Definition for binary tree
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     private TreeNode lastNode = null;
     public void flatten(TreeNode root) {
         if(root == null)
             return;

         if(lastNode != null){
             lastNode.left = null;
             lastNode.right = root;
         }

         lastNode = root;

         TreeNode right = root.right;
         flatten(root.left);
         flatten(right);
     }
 }