剑指Offer之Java算法习题精讲链表与二叉树专项训练
作者：明天一定.
跟着思路走，之后从简单题入手，反复去看，做过之后可能会忘记，之后再做一次，记不住就反复做，反复寻求思路和规律，慢慢积累就会发现质的变化
题目一
链表题——反转链表
根据单链表的头节点head来返回反转后的链表
具体题目如下
解法
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
ListNode pre,cur,nxt;
pre = null;
cur = head;
nxt = head;
while(cur!=null){
nxt = cur.next;
cur.next = pre;
pre = cur;
cur = nxt;
}
return pre;
}
}题目二
链表题——反转链表
按照一定数量的节点来进行反转并返回反转之后的链表
具体题目如下
解法
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
if (head == null) return null;
ListNode a, b;
a = b = head;
for (int i = 0; i < k; i++) {
if (b == null) return head;
b = b.next;
}
ListNode newHead = reverse(a, b);
a.next = reverseKGroup(b, k);
return newHead;
}
ListNode reverse(ListNode a, ListNode b) {
ListNode pre,cur,nxt;
pre = null;
cur = a;
nxt = a;
while(cur!=b){
nxt = cur.next;
cur.next = pre;
pre = cur;
cur = nxt;
}
return pre;
}
}题目三
链表题——回文链表
根据单链表的头节点head来判断该链表是否是回文链表，并返回结果
具体题目如下
解法：后序遍历与left比较
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
ListNode left;
public boolean isPalindrome(ListNode head) {
left = head;
return traverse(head);
}
boolean traverse(ListNode right){
if (right == null) return true;
boolean res = traverse(right.next);
res = res && (right.val == left.val);
left = left.next;
return res;
}
}题目四
二叉树题——翻转二叉树
根据所给的二叉树根节点root来翻转此二叉树，并返回翻转后的二叉树根节点
具体题目如下
解法
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
if(root==null){
return null;
}
TreeNode lf = invertTree(root.left);
TreeNode rg = invertTree(root.right);
root.left = rg;
root.right = lf;
return root;
}
}题目五
二叉树题——填充节点
给定一个完美二叉树，填充该二叉树每个节点的下一个右侧节点指针
具体题目如下
解法
/*
// Definition for a Node.
class Node {
public int val;
public Node left;
public Node right;
public Node next;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, Node _left, Node _right, Node _next) {
val = _val;
left = _left;
right = _right;
next = _next;
}
};
*/
class Solution {
public Node connect(Node root) {
if(root==null) return null;
method(root.left,root.right);
return root;
}
public void method(Node left,Node right){
if (left == null || right == null) {
return;
}
left.next = right;
method(left.left,left.right);
method(right.left,right.right);
method(left.right,right.left);
}
}题目六
二叉树链表题——将二叉树展开为链表
根据给定的二叉树根节点root，将此二叉树展开为单链表
具体题目如下
解法
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public void flatten(TreeNode root) {
if (root == null) return;
flatten(root.left);
flatten(root.right);
TreeNode left = root.left;
TreeNode right = root.right;
root.left = null;
root.right = left;
TreeNode p = root;
while (p.right != null) {
p = p.right;
}
p.right = right;
}
}到此这篇关于剑指Offer之Java算法习题精讲链表与二叉树专项训练的文章就介绍到这了,更多相关Java 链表内容请搜索脚本之家以前的文章或继续浏览下面的相关文章希望大家以后多多支持脚本之家！