二叉树展开为链表

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。

示例 1：

输入：root = [1,2,5,3,4,null,6]
输出：[1,null,2,null,3,null,4,null,5,null,6]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [0]
输出：[0]

解法一：递归

通过递归的方法，对树进行前序遍历，存入到list中，遍历list。

代码为：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    //递归
    public void flatten(TreeNode root) {
        List<TreeNode> list = new LinkedList<>();
        //前序遍历
        first(root,list);
        int len = list.size();
        //遍历list，展开为链表
        for (int i = 1;i<len;i++){
            TreeNode temp = list.get(i-1),right = list.get(i);
            temp.left = null;
            temp.right = right;
        }
    }
    public void first(TreeNode root,List<TreeNode> list){
        if (root!=null){
            list.add(root);
            first(root.left,list);
            first(root.right,list);
        }
    }
}

解法二：迭代

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    //迭代
    public void flatten(TreeNode root) {
        List<TreeNode> list = new LinkedList<>();
        Stack<TreeNode> stack = new Stack<>();
        while (root!=null || !stack.isEmpty()){
            while (root!=null){
                list.add(root);
                stack.add(root);
                root = root.left;
            }
            root = stack.pop();
            root = root.right;
        }
        int len = list.size();
        for (int i = 1;i<len;i++){
            TreeNode t = list.get(i-1),right = list.get(i);
            t.left = null;
            t.right = right;
        }
    }
}

解法三：在进行前序遍历的同时进行展开

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    //在进行前序遍历的同时进行展开
    public void flatten(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        if (root!=null) stack.push(root);
        TreeNode pre = null;
        while (!stack.isEmpty()){
            TreeNode current = stack.pop();
            if (pre!=null){
                pre.left = null;
                pre.right = current;
            }
            if (current.right!=null){
                TreeNode right = current.right;
                stack.push(right);
            }
            if (current.left!=null){
                TreeNode left = current.left;
                stack.push(left);
            }
            pre = current;
        }
    }
}

解法四：不利用前序遍历，用寻找前驱节点

​ 具体做法是，对于当前节点，如果其左子节点不为空，则在其左子树中找到最右边的节点，作为前驱节点，将当前节点的右子节点赋给前驱节点的右子节点，然后将当前节点的左子节点赋给当前节点的右子节点，并将当前节点的左子节点设为空。对当前节点处理结束后，继续处理链表中的下一个节点，直到所有节点都处理结束。

代码为：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
     //不利用前序遍历，用寻找前驱节点
    public void flatten(TreeNode root) {
        TreeNode cur = root;
        while (cur!=null){
            if (cur.left!=null){
                TreeNode left = cur.left;
                TreeNode pre = left;
                while (pre.right!=null){
                    pre = pre.right;
                }
                pre.right = cur.right;
                cur.left = null;
                cur.right = left;
            }
            cur = cur.right;
        }

    }
}