26、二叉搜索树与双向链表

一、题目

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。

二、解法

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

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

    }

}
*/
public class Solution {
    public TreeNode Convert(TreeNode pRootOfTree) {
        if(pRootOfTree == null)
            return null;
        //如果是叶子结点
        if(pRootOfTree.left == null && pRootOfTree.right == null)
            return pRootOfTree;
        //1、先从左子树遍历
        TreeNode left = Convert(pRootOfTree.left);
        TreeNode p = left;//p指向当前链表的最后一个结点
        while(p != null && p.right != null)
            p = p.right;
        //2、与当前的pRootOfTree进行连接
        //p.right = pRootOfTree; pRootOfTree.left = p;
        if(left != null){
            p.right = pRootOfTree;
            pRootOfTree.left = p;
        }
        //3、右子树转换为双向链表
        TreeNode right = Convert(pRootOfTree.right);
        if(right != null){
            pRootOfTree.right = right;
            right.left = pRootOfTree;
        }
        return left != null ? left : pRootOfTree;
    }
}