LeetCode 114. Flatten Binary Tree to Linked List

Problem Description:

Given a binary tree, flatten it to a linked list in-place.

For example, given the following tree:

    1
   / \
  2   5
 / \   \
3   4   6

The flattened tree should look like:

1
 \
  2
   \
    3
     \
      4
       \
        5
         \
          6
题解：
这个题我首先想到的是前序遍历。直接采用前序遍历会因为遍历的同时修改造成错误，需要提前记录下左右子树的引用。代码如下

TreeNode head, tail;
    public void flatten(TreeNode root) {
        head = null;
        tail = null;
        dfs(root);
        
    }
    private void dfs(TreeNode root) {
        if(root == null) return;
        // System.out.println(root.val);
        TreeNode right = root.right;
        TreeNode left = root.left;
        if(head == null) {
            head = tail = root;
        } else {
            tail.right = root;
            tail.left = null;
            tail = root;
        }
        dfs(left);
        dfs(right);
    }

　　看到讨论区有后序遍历的方法，感觉更加精简：

private TreeNode prev = null;

public void flatten(TreeNode root) {
    if (root == null)
        return;
    flatten(root.right);
    flatten(root.left);
    root.right = prev;
    root.left = null;
    prev = root;
}