Algorithm/src/main/java/tree/Flatten.java

package tree;

import java.util.ArrayList;
import java.util.Deque;
import java.util.LinkedList;
import java.util.List;

/**
 * 题目： 114. 二叉树展开为链表  (rightSideView)
 * 描述：给你二叉树的根结点 root ，请你将它展开为一个单链表：
 * 展开后的单链表应该同样使用 TreeNode ，其中 right 子指针指向链表中下一个结点，而左子指针始终为 null 。
 * 展开后的单链表应该与二叉树 先序遍历 顺序相同。

 * 链接：https://leetcode.cn/problems/flatten-binary-tree-to-linked-list/

 */
public class Flatten {
    public void inOrderTraversal(TreeNode root,List<TreeNode>list){
        if(root!=null) {
            list.add(root);
            inOrderTraversal(root.left, list);
            inOrderTraversal(root.right, list);
        }
    }
    //O(N)空间复杂度 递归前序
    public void flatten(TreeNode root) {
        List<TreeNode>list=new ArrayList<>();
        TreeNode head=new TreeNode(0);
        TreeNode tp=head;
        inOrderTraversal(root,list);
        for (TreeNode treeNode : list) {
            tp.left=null;
            tp.right=treeNode;
            tp=tp.right;
        }
    }
    //迭代前序遍历也要回
    public void flatten1(TreeNode root) {
        if (root == null) {
            return;
        }
        Deque<TreeNode> stack = new LinkedList<TreeNode>();
        stack.push(root);
        TreeNode prev = null;
        while (!stack.isEmpty()) {
            TreeNode curr = stack.pop();
            if (prev != null) {
                prev.left = null;
                prev.right = curr;
            }
            TreeNode left = curr.left, right = curr.right;
            if (right != null) {
                stack.push(right);
            }
            if (left != null) {
                stack.push(left);
            }
            prev = curr;
        }
    }
    //O(1)空间
    public void flatten2(TreeNode root) {
        TreeNode curr = root;
        while (curr != null) {
            if (curr.left != null) {
                TreeNode next = curr.left;
                TreeNode predecessor = next;
                while (predecessor.right != null) {
                    predecessor = predecessor.right;
                }
                predecessor.right = curr.right;
                curr.left = null;
                curr.right = next;
            }
            curr = curr.right;
        }
    }
}