题解 | #二叉树展开为单链表#

import java.util.*;

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

/**
 * NC332 二叉树展开为单链表
 * @author d3y1
 */
public class Solution {
    private ArrayList<TreeNode> list = new ArrayList<>();
    private TreeNode rightNode;

    /**
     * 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
     *
     * 程序入口
     *
     * @param root TreeNode类
     * @return TreeNode类
     */
    public TreeNode expandTree (TreeNode root) {
        // return solution1(root);
        // return solution2(root);
        // return solution3(root);
        return solution4(root);
    }

    /**
     * dfs
     * @param root
     * @return
     */
    private TreeNode solution1(TreeNode root){
        preOrder(root);
        rightLinkNodes();

        return root;
    }

    /**
     * 递归: 前序遍历
     * @param root
     */
    private void preOrder(TreeNode root){
        if(root == null){
            return;
        }

        list.add(root);
        preOrder(root.left);
        preOrder(root.right);
    }

    /**
     * 右链接各节点
     */
    private void rightLinkNodes(){
        int size = list.size();
        if(size > 0){
            for(int i=1; i<size; i++){
                list.get(i-1).left = null;
                list.get(i-1).right = list.get(i);
            }
            list.get(size-1).left = null;
            list.get(size-1).right = null;
        }
    }

    /**
     * 非递归: 栈
     * @param root
     * @return
     */
    private TreeNode solution2(TreeNode root){
        Stack<TreeNode> stack = new Stack<>();

        if(root == null){
            return null;
        }

        stack.push(root);

        TreeNode curr;
        TreeNode pre = null;
        while(!stack.isEmpty()){
            curr = stack.pop();
            if(curr.right != null){
                stack.push(curr.right);
            }
            if(curr.left != null){
                stack.push(curr.left);
            }
            if(pre != null){
                pre.left = null;
                pre.right = curr;
            }
            pre = curr;
        }
        pre.left = null;
        pre.right = null;

        return root;
    }

    /**
     * 循环遍历(非基本的树遍历)
     * @param root
     * @return
     */
    private TreeNode solution3(TreeNode root){
        if(root == null){
            return null;
        }

        TreeNode curr = root;
        TreeNode pre;
        while(curr != null){
            if(curr.left != null){
                pre = curr.left;
                while(pre.right != null){
                    pre = pre.right;
                }
                pre.right = curr.right;
                curr.right = curr.left;
                curr.left = null;
            }
            curr = curr.right;
        }

        return root;
    }

    /**
     * dfs
     * @param root
     * @return
     */
    private TreeNode solution4(TreeNode root){
        postOrder(root);

        return root;
    }

    /**
     * 递归: 后序遍历(变体)[右左中]
     * 从后往前处理
     * 妙!
     * @param root
     */
    private void postOrder(TreeNode root){
        if(root == null){
            return;
        }

        postOrder(root.right);
        postOrder(root.left);

        root.right = rightNode;
        root.left = null;
        rightNode = root;
    }
}