剑指Offer之Java算法习题精讲二叉树与链表剑指Offer之Java算法习题精讲二

题目一
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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

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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

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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

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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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