1、前言

2、思路

此题的思路分为三步：

• 将二叉树左子树展开为链表
• 将二叉树右子树展开为链表

• 将右子树链接到左子树的右边

  
  思路

3、代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {

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