Java 二叉树
明天一定. 人气:0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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