題目
輸入一棵二叉搜索樹,將該二叉搜索樹轉換成一個排序的雙向鏈表。要求不能創建任何新的結點,只能調整樹中結點指針的指向。
分析
二叉搜索樹轉變為排序的雙向鏈表,即將二叉搜索樹按照中序遍歷。中序遍歷的結果即是排好序的結果,將中序遍歷時的每一個節點的left和right進行調整便可實現排好序的雙向鏈表
代碼
import java.util.Stack;
public class Solution {
public static void main(String[] args) {
// 創建根節點
TreeNode root = new TreeNode(6);
Solution solution = new Solution();
int[] a = { 4, 7, 2, 5};
// 插入節點
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;
}
}
// 將二叉搜索樹轉換為排序的雙向鏈表
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();
// 處理頭節點
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;
}
// 遞歸方式中序遍歷二叉搜索樹
public void inFindDiGui(TreeNode root) {
TreeNode p = root;
if (null != p) {
inFindDiGui(p.left);
System.out.print(p.val + "->");
inFindDiGui(p.right);
}
}
// 非遞歸方式中序遍歷二叉搜索樹
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;
}
}
}
// 初始化二叉搜索樹
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;
}
}