lintcode：将二叉树拆成链表

假链表

1
\
1          2
/ \          \
2   5    =>    3
/ \   \          \
3   4   6          4
\
5
\
6

/**
* Definition of TreeNode:
* public class TreeNode {
*     public int val;
*     public TreeNode left, right;
*     public TreeNode(int val) {
*         this.val = val;
*         this.left = this.right = null;
*     }
* }
*/
public class Solution {
/**
* @param root: a TreeNode, the root of the binary tree
* @return: nothing
* 前序遍历：根左右
* 修改根的right指向左部分
* 左部分的right指向右部分
*/
public void flatten(TreeNode root) {
// write your code here
// 空结点
if(root == null)
return;
// 左右结点都空
if(root.left==null && root.right==null)
return;
// 左结点空
if(root.left==null){
TreeNode right = root.right;
flatten(right);
return;
}
// 右节点空
if(root.right==null){
TreeNode left = root.left;
root.right = left;

flatten(left);
root.left = null;// left结点设置为空
return;
}
// 左右结点都不空
TreeNode left = root.left;
TreeNode right = root.right;
root.right = left;

flatten(left);

TreeNode p = left;
while(p.right!=null){ // 找到left部分的最右结点
p = p.right;
}
p.right = right;
flatten(right);

root.left = null; // left结点设置为空
return;
}
}