lintcode-easy-Flatten Binary Tree to Linked List

Flatten a binary tree to a fake "linked list" in pre-order traversal.

Here we use the right pointer in TreeNode as the nextpointer in ListNode.

Example

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

Note

Don‘t forget to mark the left child of each node to null. Or you will get Time Limit Exceeded or Memory Limit Exceeded.

Challenge

Do it in-place without any extra memory.

递归一下，注意如果左子树为null直接flatten右子树然后返回，否则超时

/**
 * 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
     */
    public void flatten(TreeNode root) {
        // write your code here
        if(root == null)
            return;

        if(root.left == null && root.right == null)
            return;

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

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

        TreeNode temp = root.right;

        if(root.left != null){
            root.right = root.left;
            root.left = null;
            TreeNode p = root;
            while(p.right != null)
                p = p.right;

        p.right = temp;
        }

        return;
    }
}