Leetcode 二叉树 669

二叉树： 

669. Trim a Binary Search Tree

class Solution {
public:
    TreeNode* trimBST(TreeNode* root, int low, int high) {
        if(root == NULL) return NULL;
        if(root->val < low) return trimBST(root->right, low, high);
        if(root->val > high) return trimBST(root->left, low , high);

        root->left = trimBST(root->left, low, high);
        root->right = trimBST(root->right, low, high);

        return root;  
    }
};

108. Convert Sorted Array to Binary Search Tree

class Solution {
public:
    TreeNode* sortedArrayToBST(vector& nums) {
        return buildBST(nums, 0, nums.size());
    }

    TreeNode* buildBST(vector& nums, int left, int right){
        //左闭右开区间
        if(left >= right) return NULL;
        int mid = left + (right-left)/2;
        TreeNode* newNode = new TreeNode(nums[mid]);
        newNode->left = buildBST(nums, left, mid);
        newNode->right = buildBST(nums, mid+1, right);
        return newNode;
    }
};

确定好区间的开闭

538. Convert BST to Greater Tree

class Solution {
public:
    int sum = 0;
    TreeNode* convertBST(TreeNode* root) {
        if(root == NULL) return NULL;
        root->right = convertBST(root->right);
        sum += root->val;
        root->val = sum;
        root->left = convertBST(root->left);
        return root;
    }
};

Top interview 150(String):

88. Merge Sorted Array

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

因为这道题要直接加在nums1里面，所以从后往前

（需不需要resize nums1）

27. Remove Element

class Solution {
public:
    int removeElement(vector& nums, int val) {
        int index = 0;
        for(int i=0; i

26. Remove Duplicates from Sorted Array

1.

lass Solution {
public:
    int removeDuplicates(vector& nums) {
        int slow = 1, fast = 1;
        while(fast < nums.size()){
            if(nums[fast] == nums[fast-1]){
                fast++;
            }else{
                nums[slow] = nums[fast];
                slow++;
                fast++;
            }
        }
        return slow;
    }
};

2.

class Solution {
public:
    int removeDuplicates(vector& nums) {
        int slow = 0, fast = 0;
        while(fast < nums.size()){
            if(nums[fast] != nums[slow]){
                slow++;
                nums[slow] = nums[fast];
            }
            fast++;
        }
        return slow+1;
    }
};

要注意如何保留下来第一个数字

80. Remove Duplicates from Sorted Array II

class Solution {
public:
    int removeDuplicates(vector& nums) {
        int dupnum = 0;
        int slow = 0, fast = 0;
        while(fast < nums.size()){
            if(nums[fast] == nums[slow]){
                if(slow < fast && dupnum <= 1){
                    slow++;
                    nums[slow] = nums[fast];
                }
            }else{
                dupnum = 0;
                slow++;
                nums[slow] = nums[fast];
                
                
            }
            fast++;
            dupnum++;
        } 
        return slow+1;
    }
};

出现的一些错误：

1.记录重复次数的当出现>2个的情况时，不需要将dupnum赋值0, 如果赋值0,后面又会重新计数，加进来， 遇到不同的在改为0就可以了

2.要添加一个slow