If you notice carefully in the flattened tree, each node‘s right child points to the next node of a pre-order traversal.

原题链接：https://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/

题目：给定二叉树，按前序位置展平成一个链表。

思路：递归处理。把右子树放到左子树之后，并清空左子树。

	public void flatten(TreeNode root) {
		if(root == null)
			return;
		flatten(root.left);
		flatten(root.right);
		TreeNode tmp = root;
		if(tmp.left == null)
			return;
		else
			tmp = tmp.left;
		while(tmp.right != null)
			tmp = tmp.right;
		tmp.right = root.right;
		root.right = root.left;
		root.left = null;
	}
    // Definition for binary tree
    public class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;
        TreeNode(int x) { val = x; }
    }

	public void flatten(TreeNode root){
		while(root != null){
			if(root.left != null){
				TreeNode tmp = root.left;
				while(tmp.right != null)
					tmp = tmp.right;
				tmp.right = root.right;
				root.right = root.left;
				root.left = null;
			}
			root = root.right;
		}
	}