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劍指Offer之Java算法習(xí)題精講鏈表與二叉樹(shù)專項(xiàng)訓(xùn)練

更新時(shí)間：2022年03月22日 08:38:17 作者：明天一定.

跟著思路走，之后從簡(jiǎn)單題入手，反復(fù)去看，做過(guò)之后可能會(huì)忘記，之后再做一次，記不住就反復(fù)做，反復(fù)尋求思路和規(guī)律，慢慢積累就會(huì)發(fā)現(xiàn)質(zhì)的變化

題目一

鏈表題——反轉(zhuǎn)鏈表

根據(jù)單鏈表的頭節(jié)點(diǎn)head來(lái)返回反轉(zhuǎn)后的鏈表

具體題目如下

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;
    }
}

題目二

鏈表題——反轉(zhuǎn)鏈表

按照一定數(shù)量的節(jié)點(diǎn)來(lái)進(jìn)行反轉(zhuǎn)并返回反轉(zhuǎn)之后的鏈表

具體題目如下

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;
    }
}

題目三

鏈表題——回文鏈表

根據(jù)單鏈表的頭節(jié)點(diǎn)head來(lái)判斷該鏈表是否是回文鏈表，并返回結(jié)果

具體題目如下

解法：后序遍歷與left比較

/**
 * 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;
    }
}

題目四

二叉樹(shù)題——翻轉(zhuǎn)二叉樹(shù)

根據(jù)所給的二叉樹(shù)根節(jié)點(diǎn)root來(lái)翻轉(zhuǎn)此二叉樹(shù)，并返回翻轉(zhuǎn)后的二叉樹(shù)根節(jié)點(diǎn)

具體題目如下

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;
    }
}

題目五

二叉樹(shù)題——填充節(jié)點(diǎn)

給定一個(gè)完美二叉樹(shù)，填充該二叉樹(shù)每個(gè)節(jié)點(diǎn)的下一個(gè)右側(cè)節(jié)點(diǎn)指針

具體題目如下

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);
    }
}

題目六

二叉樹(shù)鏈表題——將二叉樹(shù)展開(kāi)為鏈表

根據(jù)給定的二叉樹(shù)根節(jié)點(diǎn)root，將此二叉樹(shù)展開(kāi)為單鏈表

具體題目如下

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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