二叉搜索树与双向链表

题目描述

输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点，只能调整树中结点指针的指向。

双向链表：它的每个数据结点中都有两个指针，分别指向直接后继和直接前驱

Solution 1:

public class Solution {

    TreeNode realhead = null; //realhead 是双向链表的头

    TreeNode cur = null;

    public TreeNode Convert(TreeNode pRootOfTree) {

        convertSub(pRootOfTree);

        return realhead;

    }

    

    private void convertSub(TreeNode pRootOfTree) {

        if (pRootOfTree == null)

            return;

        convertSub(pRootOfTree.left);

        if (realhead == null) {

            cur = pRootOfTree; //把最左的节点赋给cur和realhead，只在最开始链表没数据时会进入这个if语                                                   句

            realhead = pRootOfTree;

        } else {

            cur.right = pRootOfTree; //双向链表，所以必须得对左右声明赋值

            pRootOfTree.left = cur;

            cur = pRootOfTree; //把该节点设成新的head

        }

        convertSub(pRootOfTree.right);

    }

}

Solution 2: //借助stack    

public TreeNode Convert(TreeNode root) {
    if (root == null)
        return null;
    Stack<TreeNode> stack = new Stack<>();
    TreeNode realHead = null; //用于保存中序遍历的上一节点
    TreeNode cur = null; //用于保存中序遍历的上一节点
    boolean isFirst = true;
    stack.push(root);
    while (!stack.isEmpty()) {
        TreeNode p = stack.peek();
        while (p != null) {
            p = p.left;
            stack.push(p);
        }
        stack.pop();
        if (!stack.isEmpty()) {
            p = stack.pop();
            if (isFirst) {//只在最开始链表没数据时会进入这个if语句
                realHead = p;
                cur = p;
                isFirst = false;
            } else {
                cur.right = p;
                p.left = cur;
                cur = p;
            }
            stack.push(p.right);
        }
    }
    return root;
}