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。
- 主要思想:递归
- 题目要点: 根据二叉搜索树的中序遍历即是有序数组的特性,对二叉搜素树进行中序遍历,在遍历过程中调整节点指针的指向。
/**
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
*/
public class Solution {
TreeNode head = null;
TreeNode realHead = null;
public TreeNode Convert(TreeNode pRootOfTree) {
helper(pRootOfTree);
return realHead;
}
private void helper(TreeNode pRootOfTree) {
if (pRootOfTree == null)
return;
helper(pRootOfTree.left);
if (head == null) {
head = pRootOfTree;
realHead = pRootOfTree;
} else {
//先调整head与pRootOfTree的指向
pRootOfTree.left = head;
head.right = pRootOfTree;
//令head指向pRootOfTree,用于下一轮循环
head = pRootOfTree;
}
helper(pRootOfTree.right);
}
}