#83 Remove Duplicates from Sorted List

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* deleteDuplicates(struct ListNode* head) {
    struct ListNode *p,*temp;
    if (head) 
    {
        p = head;
        while (p->next) 
        {
            if (p->val != p->next->val) 
                p = p->next;
            else
            {
                temp = p->next;
                p->next = p->next->next;
                free(temp);
            }
        }
    }
    return head;
}

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:
You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1and nums2 are m and n respectively.

//0ms
void merge(int* nums1, int m, int* nums2, int n) {
    int index = m + n -1, i = m - 1, j = n - 1;
    while(j>=0)
    {
        if(i < 0 || nums1[i] < nums2[j])
            nums1[index--] = nums2[j--];
        else
            nums1[index--] = nums1[i--];
    }
}

Given two binary trees, write a function to check if they are equal or not.

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

//0ms
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
bool isSameTree(struct TreeNode* p, struct TreeNode* q) {
    if(p==NULL || q==NULL)
        return p == q;
    if(p->val == q->val)
        return isSameTree(p->left,q->left) && isSameTree(p->right,q->right);
    else
        return false;
}

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1
   /   2   2
 / \ / 3  4 4  3

But the following is not:

    1
   /   2   2
   \      3    3

//4ms
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
 
bool dfs(struct TreeNode* root1, struct TreeNode *root2)
{
    if(root1 == NULL||root2 == NULL)
        return root1 == root2;
    if(root1->val != root2->val )
        return false;
    else
        return dfs(root1->left,root2->right) && dfs(root1->right,root2->left);
}

bool isSymmetric(struct TreeNode* root) {
    if(!root || (!root->left && !root->right))//空树||仅仅有根结点
        return true;
    else
        return dfs(root->left,root->right);
}