题解 | #牛群的树形结构展开II#

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param root TreeNode类
     * @return TreeNode类
     */
    public static LinkedList<Integer> linkedList = new LinkedList<>();
    public TreeNode flattenII (TreeNode root) {
        // write code here
        search(root);
        TreeNode result = new TreeNode(0);
        TreeNode node = result;
        for (int i = 0; i < linkedList.size(); i++) {
            result.right = new TreeNode(linkedList.get(i));
            result.left = null;
            result = result.right;
        }
        result.right = null;
        result.left = null;
        return node.right;
    }

    public void search(TreeNode root) {
        if (root == null) {
            return;
        }
        search(root.left);
        linkedList.add(root.val);
        search(root.right);
    }

}

本题考察的知识点是二叉树的中序遍历和链表，所用编程语言是java。

二叉树的中序遍历是先遍历左子树，再遍历分支节点，最后遍历右子树。我先用LinkedList存储了二叉树的中序遍历序列，然后重建链表。