LeetCode

题目：

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.

思路：

递归，先flatten左边的，再flatten右边的，然后进行连接

package tree;

public class FlattenBinaryTreeToLinkedList {

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

}