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。
分析
二叉搜索树转变为排序的双向链表,即将二叉搜索树按照中序遍历。中序遍历的结果即是排好序的结果,将中序遍历时的每一个节点的left和right进行调整便可实现排好序的双向链表
代码
import java.util.Stack;
public class Solution {
public static void main(String[] args) {
// 创建根节点
TreeNode root = new TreeNode(6);
Solution solution = new Solution();
int[] a = { 4, 7, 2, 5};
// 插入节点
for (int i = 0; i < a.length; i++) {
solution.initTree(root, a[i]);
}
solution.inFindDiGui(root);
System.out.println();
solution.inFindNoDiGui(root);
System.out.println();
TreeNode head = solution.Convert(root);
TreeNode node = head;
while(node != null){
System.out.print(node.val+"->");
node = node.right;
}
}
// 将二叉搜索树转换为排序的双向链表
public TreeNode Convert(TreeNode pRootOfTree) {
TreeNode p = pRootOfTree;
Stack<TreeNode> stack = new Stack<TreeNode>();
boolean flag = true;
TreeNode head = null;
TreeNode pre = null;
while(p!=null || !stack.isEmpty()){
while(p!=null){
stack.push(p);
p = p.left;
}
if(!stack.isEmpty()){
p = stack.pop();
// 处理头节点
if(flag){
p.left = null;
head = p;
pre = p;
flag = false;
} else {
pre.right = p;
p.left = pre;
pre = pre.right;
}
p = p.right;
}
}
return head;
}
// 递归方式中序遍历二叉搜索树
public void inFindDiGui(TreeNode root) {
TreeNode p = root;
if (null != p) {
inFindDiGui(p.left);
System.out.print(p.val + "->");
inFindDiGui(p.right);
}
}
// 非递归方式中序遍历二叉搜索树
public void inFindNoDiGui(TreeNode root) {
TreeNode p = root;
Stack<TreeNode> stack = new Stack<TreeNode>();
while(p!=null || !stack.isEmpty()){
while(p!=null){
stack.push(p);
p = p.left;
}
if(!stack.isEmpty()){
p = stack.pop();
System.out.print(p.val + "->");
p = p.right;
}
}
}
// 初始化二叉搜索树
public void initTree(TreeNode root, int val) {
if (root != null) {
if (val < root.val) {
if (root.left == null) {
root.left = new TreeNode(val);
} else {
initTree(root.left, val);
}
} else {
if (root.right == null) {
root.right = new TreeNode(val);
} else {
initTree(root.right, val);
}
}
}
}
}
class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}