剑指Offer之Java算法习题精讲链表与二叉树专项训练

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;
    }
}

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