【一次过】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。

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解题思路：

与Lintcode 378. 将二叉查找树转换成双链表类似，不同的是这个不需要新建节点。

/**
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;
        
        TreeNode left = Convert(pRootOfTree.left);
        TreeNode right = Convert(pRootOfTree.right);
        
        if(left != null){
            TreeNode leftCur = left;
                while(leftCur.right != null)
            leftCur = leftCur.right;
            
            pRootOfTree.left = leftCur;
            leftCur.right = pRootOfTree;
        }
        
        if(right != null){
            pRootOfTree.right = right;
            right.left = pRootOfTree;
        }
        
        
        while(pRootOfTree.left != null)
            pRootOfTree = pRootOfTree.left;
        
        return pRootOfTree;
    }
}