概目描述
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。
输入输出示例：
{10,6,4,8,14,12,16}
from left to right:4,6,8,10,12,14,16 from right to left:16,14,12,10,8,6,4
{1,2,3,4,5}
from left to right:1,2,3,4,5 from right to left:5,4,3,2,1
{1}
from left to right:1 from right to left:1
分析
方法一：可用二叉树中序遍历方法来遍历BST，遍历过程中，转接结点的左右结点。示例解析图如下。
方法二：用递归法来求解。把左子树排成双向链表后，返回第一个结点，再把root结点接到左表后面，把右子树也排成双向链表并返回第一个结点，把右表接到root结点后面。
代码一
import java.util.Stack;
public class BST2List {
public static void main(String[] args) {
// TODO Auto-generated method stub
TreeNode root = new TreeNode(10);
root.left = new TreeNode(6);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(8);
root.right = new TreeNode(14);
root.right.left = new TreeNode(12);
root.right.right = new TreeNode(16);
root = convert(root);
while(root != null){
System.out.println(root.value);
root = root.right;
}
}
public static TreeNode convert(TreeNode root){
Stack<TreeNode> stack = new Stack<>();
if(root == null){
return null;
}
stack.push(root);
TreeNode sNode = root;
while(sNode.left != null){
stack.push(sNode.left);
sNode = sNode.left;
}
TreeNode node1 = stack.pop();
root = node1;
if(stack.isEmpty() && root.right!=null){
stack.push(root.right);
}
while(!stack.isEmpty()){
TreeNode node2 = stack.pop();
node1.right = node2;
node2.left = node1;
node1 = node2;
TreeNode tempNode = node2.right;
if(tempNode != null){
stack.push(tempNode);
while(tempNode.left != null){
stack.push(tempNode.left);
tempNode = tempNode.left;
}
}
}
return root;
}
}
代码二
public class BST2List2 {
public static void main(String[] args) {
// TODO Auto-generated method stub
TreeNode root = new TreeNode(10);
root.left = new TreeNode(6);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(8);
root.right = new TreeNode(14);
root.right.left = new TreeNode(12);
root.right.right = new TreeNode(16);
root = convert(root);
while(root != null){
System.out.println(root.value);
root = root.right;
}
}
public static TreeNode convert(TreeNode root){
if(root == null){
return null;
}
if(root.left==null && root.right==null){
return root;
}
TreeNode left = convert(root.left);
TreeNode p = left;
while(p!=null && p.right != null){
p=p.right;
}
if(left != null){
p.right=root;
root.left=p;
}
TreeNode right = convert(root.right);
if(right != null){
right.left = root;
root.right = right;
}
return left != null ? left : root;
}
}
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作者：另一个我竟然存在
来源：CSDN
原文：blog.csdn.net/qq_24034545…
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