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 next pointer in ListNode.

For example,
Given

         1
        / \
       2   5
      / \   \
     3   4   6

The flattened tree should look like:

   1
    \
     2
      \
       3
        \
         4
          \
           5
            \
             6

分析：

把所有的树节点按照PRE-ORDER的顺序存在ArrayList里面，然后遍历ArrayList里的节点，调整它们的left and right child指针。

 /**
  * 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
      * cnblogs.com/beiyeqingteng
      */
     public void flatten(TreeNode root) {
         // write your code here
         ArrayList<TreeNode> list = new ArrayList<TreeNode>();
         preOrder(root, list);

         for (int i = ; i < list.size(); i++) {
             list.get(i).left = null;
             if (i == list.size() - ) {
                 list.get(i).right = null;
             } else {
                 list.get(i).right = list.get(i + );
             }
         }
     }

     public void preOrder(TreeNode root, ArrayList<TreeNode> list) {
         if (root != null) {
             list.add(root);
             preOrder(root.left, list);
             preOrder(root.right, list);
         }
     }
 }

自己后来写出的方法：

 public class Solution {

     public void flatten(TreeNode root) {
         helper(root);
     }

     public TreeNode helper(TreeNode root) {
         if (root == null) return root;
         TreeNode leftHead = helper(root.left);
         TreeNode rightHead = helper(root.right);
         root.left = null;
         if (leftHead != null) {
             root.right = leftHead;
             TreeNode current = root;
             while (current.right != null) {
                 current = current.right;
             }
             current.right = rightHead;
         }
         return root;
     }
 }

还有一种是递归的方式，下面的方法参考了另一个网友的做法，觉得这种方法非常容易理解，而且具有一般性。

解法2：递归构建

假设某节点的左右子树T(root->left)和T(root->right)已经flatten成linked list了：

1
  /    \
 2     5
  \       \
   3      6 <- rightTail
     \
      4  <- leftTail

如何将root、T(root->left)、T(root->right) flatten成一整个linked list？显而易见：

temp = root->right
root->right  = root->left
root->left = NULL
leftTail->right = temp

这里需要用到已经flatten的linked list的尾部元素leftTail, rightTail。所以递归返回值应该是生成的整个linked list的尾部。

 class Solution {
     public void flatten(TreeNode root) {
         flattenBT(root);
     }

     TreeNode flattenBT(TreeNode root) {
         if(root == null) return null;
         TreeNode leftTail = flattenBT(root.left);
         TreeNode rightTail = flattenBT(root.right);
10         if(root.left != null) {
             TreeNode temp = root.right;
             root.right = root.left;
             root.left = null;
             leftTail.right = temp;
         }

17         if(rightTail != null) return rightTail;
18         if(leftTail != null) return leftTail;
19         return root;
     }
 }

需注意几个细节
ln 10：只有当左子树存在时才将它插入右子树中
ln 17-19：返回尾部元素时，需要特殊处理 (1) 有右子树的情况，(2) 无右子树但有左子树的情况，(3)左右子树均不存在的情况。

Reference:

http://bangbingsyb.blogspot.com/2014/11/leetcode-flatten-binary-tree-to-linked.html