83. Remove Duplicates from Sorted List

从排序列表中删除重复项

Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well.

Example 1:

Input: head = [1,1,2]
Output: [1,2]

Example 2:

Input: head = [1,1,2,3,3]
Output: [1,2,3]

leetcode代码

public ListNode deleteDuplicates(ListNode head) {
        if(head == null || head.next == null)return head;
        head.next = deleteDuplicates(head.next);
        return head.val == head.next.val ? head.next : head;
}

88. Merge Sorted Array

合并排序数组
You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.
Merge nums1 and nums2 into a single array sorted in non-decreasing order.
The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.

Example 1:

Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
Output: [1,2,2,3,5,6]
Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.

Example 2:

Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]
Explanation: The arrays we are merging are [1] and [].
The result of the merge is [1].

Example 3:

Input: nums1 = [0], m = 0, nums2 = [1], n = 1
Output: [1]
Explanation: The arrays we are merging are [] and [1].
The result of the merge is [1].
Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.

leetcode代码

public void merge(int[] nums1, int m, int[] nums2, int n) {
        int p1 = m-1 , p2 = n-1 ,i = m+n-1;
        while(p2 >=0 ){
            if(p1 >=0 && nums1[p1] > nums2[p2]){
                nums1[i--] = nums1[p1--];
            }
            else{
                nums1[i--] = nums2[p2--];
            }
        }
       }

94. Binary Tree Inorder Traversal

二叉树中序遍历

Given the root of a binary tree, return the inorder traversal of its nodes’ values.

Example 1:

Input: root = [1,null,2,3]
Output: [1,3,2]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

leetcode代码

/**
 * 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 List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();

        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode cur = root;

        while(cur!=null || !stack.empty()){
            while(cur!=null){
                stack.add(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            list.add(cur.val);
            cur = cur.right;
        }

        return list; 
        }
}

100. Same Tree

同一棵树

Given the roots of two binary trees p and q, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.

Example 1:

Input: p = [1,2,3], q = [1,2,3]
Output: true

Example 2:

Input: p = [1,2], q = [1,null,2]
Output: false

Example 3:

Input: p = [1,2,1], q = [1,1,2]
Output: false

leetcode代码

/**
 * 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 boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null && q == null) return true;
        return p != null && q != null && p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);     
    }
}

101. Symmetric Tree

对称树

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

Example 1:

Input: root = [1,2,2,3,4,4,3]
Output: true

Example 2:

Input: root = [1,2,2,null,3,null,3]
Output: false

leetcode代码

/**
 * 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 boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root.left);
        stack.push(root.right);
        while (!stack.empty()) {
            TreeNode n1 = stack.pop(), n2 = stack.pop();
            if (n1 == null && n2 == null) continue;
            if (n1 == null || n2 == null || n1.val != n2.val) return false;
            stack.push(n1.left);
            stack.push(n2.right);
            stack.push(n1.right);
            stack.push(n2.left);
        }
        return true;
    }
}