剑指Offer之Java算法习题精讲二叉树与链表

题目一

 解法

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
  public boolean isBalanced(TreeNode root) {
      if(root==null){
          return true;
      }else{
          return Math.abs(method(root.left) - method(root.right)) <= 1&&isBalanced(root.left) && isBalanced(root.right);
      }
  }
  public int method(TreeNode root){
      if(root==null){
          return 0;
      }else{
          return Math.max(method(root.left),method(root.right))+1;
      }
  }
}

 

题目二

 解法

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
  public boolean isSymmetric(TreeNode root) {
      if(root==null) return true;
      return method(root.left,root.right);
  }
  public boolean method(TreeNode l,TreeNode r){
      if(l==null&&r==null) return true;
      if(l==null||r==null||l.val!=r.val) return false;
      return method(l.left,r.right)&&method(l.right,r.left);
  }
}

 

题目三

 解法

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
class Solution {
  public ListNode deleteNode(ListNode head, int val) {
      ListNode temp = new ListNode(-1);
      temp.next = head;
      ListNode ans = temp;
      while(temp.next!=null){
          if(temp.next.val==val){
              temp.next = temp.next.next;
          }else{
              temp = temp.next;
          }
      }
      return ans.next;
  }
}

 

题目四

 解法

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
class Solution {
  public ListNode reverseList(ListNode head) {
      ListNode prev = null;   
      ListNode curr = head;
      while (curr != null) {
          ListNode next = curr.next;
          curr.next = prev;
          prev = curr;
          curr = next;
      }
      return prev;
  }
}

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原文链接：https://blog.csdn.net/wai_58934/article/details/123449047

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