LeetCode 114. Flatten Binary Tree to Linked List（摊平二叉树）

原题网址：https://leetcode.com/problems/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

click to show hints.

Hints:

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

方法一：递归函数将链表的尾部返回。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private TreeNode preorder(TreeNode root) {
        if (root == null) return null;
        TreeNode left = root.left;
        TreeNode right = root.right;
        if (left == null && right == null) return root;
        if (left != null) {
            TreeNode leftTail = preorder(left);
            root.left = null;
            root.right = left;
            if (right == null) return leftTail;
            leftTail.right = right;
            TreeNode rightTail = preorder(right);
            return rightTail;
        }
        TreeNode rightTail = preorder(right);
        return rightTail;
    }
    public void flatten(TreeNode root) {
        preorder(root);
    }
}

方法二：通过prev保持顺序。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private TreeNode prev;
    private void traverse(TreeNode root) {
        if (prev != null) prev.right = root;
        prev = root;
        TreeNode right = root.right;
        if (root.left != null) {
            traverse(root.left);
            root.left = null;
        }
        if (right != null) {
            traverse(right);
        }
    }
    
    public void flatten(TreeNode root) {
        if (root == null) return;
        prev = null;
        traverse(root);
    }
}

方法三：prev顺序变量放在递归参数中。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private TreeNode traverse(TreeNode prev, TreeNode root) {
        if (prev != null) prev.right = root;
        prev = root;
        TreeNode right = root.right;
        if (root.left != null) {
            prev = traverse(prev, root.left);
            root.left = null;
        }
        if (right != null) {
            prev = traverse(prev, right);
        }
        return prev;
    }
    
    public void flatten(TreeNode root) {
        if (root == null) return;
        traverse(null, root);
    }
}