114. Flatten Binary Tree to Linked List

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

solution 1 只用返回左子树和右子树的last节点而不用返回它们的first节点，因为两个子树的first节点可以用root.left和root.right直接访问到。

solution 2中因为在flatten左子树的时候会改变root.right的值，所以要预先记录一下。

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

// solution 1 : Divide & Conquer
class Solution {
    public void flatten(TreeNode root) {
         helper(root);
    }

    private TreeNode helper(TreeNode root) {
        if (root == null) {
            return null;
        }
        TreeNode leftLast = helper(root.left);
        TreeNode rightLast = helper(root.right);
        if (leftLast != null) {
            leftLast.right = root.right;
            root.right = root.left;
            root.left = null;
        }
        if (rightLast != null) {
            return rightLast;
        }
        if (leftLast != null) {
            return leftLast;
        }
        return root;
    }
}

// solution 2: traversal
class Solution {
    private TreeNode lastNode;
    public void flatten(TreeNode root) {
        if (root == null) {
            return;
        }
        if (lastNode != null) {
            lastNode.right = root;
            lastNode.left = null;
        }
        lastNode = root;
        TreeNode right = root.right;
        flatten(root.left);
        flatten(right);
    }
}