Question:

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

Hint:
If you notice carefully in the flattened tree, each node’s right child points to the next node of a pre-order traversal.

Thinking:

It’s a simple dfs question and it’s pre-order. So we need to make the left-child become right-child of root and make right-child become whole left-child’s right-child.

Solution:

public class Solution {
    public void flatten(TreeNode root) {
        if (root == null)
            return;
        dfs(root);
    }

    public TreeNode dfs(TreeNode root){

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

        TreeNode left = null;
        TreeNode right = null;
        TreeNode preleft = root.left;
        TreeNode preright = root.right;
        if (preleft == null){
            return dfs(preright);
        }
        else if(preright == null){
            root.left = null;
            root.right = preleft;
            return dfs(preleft);
        }
        else{
            left = dfs(preleft);
            right = dfs(preright);
            root.left = null;
            root.right = preleft;
            left.right = preright;
            return right;
        }
    }
}