[算法]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。

思路

因为是二叉搜索树，所以中序遍历就是有序的，这里采用非递归中序遍历，并通过前驱指针pre，进行转换。

代码

/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/

import java.util.Stack;
public class Solution {
    public TreeNode Convert(TreeNode pRootOfTree) {
        if (pRootOfTree == null) {
            return null;
        }
        Stack<TreeNode> stack = new Stack<>();
        TreeNode pre = null;//用来标记上一个节点
        TreeNode root = null;//用来返回头结点
        boolean isFirst = true;
        while (!stack.isEmpty() || pRootOfTree != null) {
            if (pRootOfTree != null) {
                stack.push(pRootOfTree);
                pRootOfTree = pRootOfTree.left;
            } else {
                pRootOfTree = stack.pop();
                if(isFirst){
                    root = pRootOfTree;
                    pre = pRootOfTree;
                    isFirst = false;
                }else{
                    pre.right = pRootOfTree;
                    pRootOfTree.left = pre;
                    pre = pRootOfTree;
                }
                pRootOfTree = pRootOfTree.right;
            }
        }
        return root;
    }
}