leetcode 563 - 653

563. Binary Tree Tilt

Input:
         1
       /   \
      2     3
Output: 1
Explanation:
Tilt of node 2 : 0
Tilt of node 3 : 0
Tilt of node 1 : |2-3| = 1
Tilt of binary tree : 0 + 0 + 1 = 1

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int findTilt(TreeNode* root) {
        if(root == NULL) return ;

        int res = ;

        postorder(root, res);

        return res;
    }
private:
    int postorder(TreeNode* root, int& res){
        if(root == NULL) return ;

        int leftsum= postorder(root->left,res);

        int rightsum = postorder(root->right,res);

        res += abs(leftsum - rightsum);

        return leftsum + rightsum + root->val;

    }
};  

566. Reshape the Matrix

class Solution {
public:
    vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
        int m = nums.size(), n = nums[].size(), o = m * n;
        if (r * c != o) return nums;
        vector<vector<int>> res(r, vector<int>(c, ));
        for (int i = ; i < o; i++) res[i / c][i % c] = nums[i / n][i % n];
        return res;
    }
};

572. Subtree of Another Tree

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    vector<TreeNode*> nodes;

public:
    bool isSubtree(TreeNode* s, TreeNode* t) {
        if (!s && !t) return true;
        if (!s || !t) return false;

        getDepth(s, getDepth(t, -));

        for (TreeNode* n: nodes)
            if (identical(n, t))
                return true;

        return false;
    }

    int getDepth(TreeNode* r, int d) {
        if (!r)
            return -;

        int depth = max(getDepth(r->left, d), getDepth(r->right, d)) + ;

        // Check if depth equals required value
        // Require depth is -1 for tree t (only return the depth, no push)
        if (depth == d)  //递归回去得时候，从树下开始记达到深度位d时记下
            nodes.push_back(r);

        return depth;
    }

    bool identical(TreeNode* a, TreeNode* b) {
        if (!a && !b) return true;
        if (!a || !b || a->val != b->val) return false;

        return identical(a->left, b->left) && identical(a->right, b->right);
    }
};

581. Shortest Unsorted Continuous Subarray

Input: [2, 6, 4, 8, 10, 9, 15]
Output: 5
Explanation: You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order.

class Solution {
    public int findUnsortedSubarray(int[] A) {
    int n = A.length, beg = -1, end = -2, min = A[n-1], max = A[0];
    for (int i=1;i<n;i++) {
      max = Math.max(max, A[i]);
      min = Math.min(min, A[n-1-i]);
      if (A[i] < max) end = i;
      if (A[n-1-i] > min) beg = n-1-i;
    }
    return end - beg + 1;
}
}

594. Longest Harmonious Subsequence

Input: [1,3,2,2,5,2,3,7]
Output: 5
Explanation: The longest harmonious subsequence is [3,2,2,2,3].

int findLHS(vector<int>& nums) {
        unordered_map<int,int>m;
        for(auto i: nums)
            m[i]++;
        int res = 0;
        for(auto it:m)
            if(m.count(it.first-1)>0)
                res = max(res, it.second+m[it.first-1]);
        return res;
    }

///////////////////////
int findLHS(vector<int>& nums) {
        unordered_map<int,int>m;
        int res = 0;
        for(auto i: nums){
            m[i]++;
            if(m.count(i+1))
                res = max(res, m[i] + m[i+1]);
            if(m.count(i-1))
                res = max(res, m[i] + m[i-1]);
        }
        return res;
    }

///////////////////////////////////////
 int findLHS(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        int len = 0;
        for(int i = 1, start = 0, new_start = 0; i<nums.size(); i++)
        {

            if (nums[i] - nums[start] > 1)
                start = new_start;
            if (nums[i] != nums[i-1])
                new_start = i;
            if(nums[i] - nums[start] == 1)
                len = max(len, i-start+1);
        }
        return len;

599. Minimum Index Sum of Two Lists

Input:
["Shogun", "Tapioca Express", "Burger King", "KFC"]
["KFC", "Shogun", "Burger King"]
Output: ["Shogun"]
Explanation: The restaurant they both like and have the least index sum is "Shogun" with index sum 1 (0+1).

  vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        vector<string>res;
        unordered_map<string,int>m;
        int min = INT_MAX;
        for(int i = ; i < list1.size(); i++) m[list1[i]] = i;
        for(int i = ; i < list2.size(); i++)
            if(m.count(list2[i]) != )
                if(m[list2[i]] + i < min) min = m[list2[i]] + i, res.clear(), res.push_back(list2[i]);
                else if(m[list2[i]] + i == min) res.push_back(list2[i]);
        return res;
    }

606. Construct String from Binary Tree

Input: Binary tree: [1,2,3,4]
       1
     /   \
    2     3
   /
  4     

Output: "1(2(4))(3)"

Explanation: Originallay it needs to be "1(2(4)())(3()())", 
but you need to omit all the unnecessary empty parenthesis pairs. 
And it will be "1(2(4))(3)".

class Solution {
public:
    string tree2str(TreeNode* t) {
        return !t ? "" : to_string(t->val) + (t->left ? "(" + tree2str(t->left) + ")" : t->right ? "()" : "")
                                           + (t->right ? "(" + tree2str(t->right) + ")" : "");
    }
};

617. Merge Two Binary Trees

Input:
	Tree 1                     Tree 2
          1                         2
         / \                       / \
        3   2                     1   3
       /                           \   \
      5                             4   7
Output:
Merged tree:
	     3
	    / \
	   4   5
	  / \   \
	 5   4   7

class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if (!t1 || !t2) return t1 ? t1 : t2;

        TreeNode* node = new TreeNode(t1->val + t2->val);
        node->left = mergeTrees(t1->left, t2->left);
        node->right = mergeTrees(t1->right, t2->right);
        return node;
    }
};

633. Sum of Square Numbers

Input: 5
Output: True
Explanation: 1 * 1 + 2 * 2 = 5

bool judgeSquareSum(int c) {
        for(int i=;i<=sqrt(c);i++) {
            int t=sqrt(c-i*i);
            if(t*t==c-i*i) return true;
        }
        return false;
    }

637. Average of Levels in Binary Tree

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
       vector<double> averageOfLevels(TreeNode* root) {
        vector<double> res;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()) {
            long temp=;
            int s=q.size();
            for(int i=;i<s;i++) {
                TreeNode* t=q.front();
                q.pop();
                if(t->left) q.push(t->left);
                if(t->right) q.push(t->right);
                temp+=t->val;
            }
            res.push_back((double)temp/s);
        }
        return res;
    }
};

645. Set Mismatch

Input: nums = [1,2,2,4]
Output: [2,3]

思路：使数组按123。。。n来排序，如果当前的内容和index不匹配，则交换

vector<int> findErrorNums(vector<int>& nums) {
        for(int i = ; i<nums.size(); i++){
            while(nums[i] != nums[nums[i] - ])swap(nums[i], nums[nums[i] - ]);
        }
        for(int i = ; i<nums.size() ; i++){
            if(nums[i] != i + )return {nums[i], i + };
        }
    }

653. Two Sum IV - Input is a BST

Input:
    5
   / \
  3   6
 / \   \
2   4   7

Target = 9

Output: True

//方法一：使用map来进行一次遍历
bool findTarget(TreeNode* root, int k) {
        unordered_set<int> set;
        return dfs(root, set, k);
    }

    bool dfs(TreeNode* root, unordered_set<int>& set, int k){
        if(root == NULL)return false;
        if(set.count(k - root->val))return true;
        set.insert(root->val);
        return dfs(root->left, set, k) || dfs(root->right, set, k);
    }

/////////////////////////////////////////////////////
//方法二：先使用中序遍历使其按小到大来排序，然后对排序后的数组进行首尾相加
bool findTarget(TreeNode* root, int k) {
        vector<int> nums;
        inorder(root, nums);
        for(int i = , j = nums.size()-; i<j;){
            if(nums[i] + nums[j] == k)return true;
            (nums[i] + nums[j] < k)? i++ : j--;
        }
        return false;
    }

    void inorder(TreeNode* root, vector<int>& nums){
        if(root == NULL)return;
        inorder(root->left, nums);
        nums.push_back(root->val);
        inorder(root->right, nums);
    }

///////////////////////////////////////////////////////

 bool findTarget(TreeNode* root, int k) {
        return dfs(root, root,  k);
    }

    bool dfs(TreeNode* root,  TreeNode* cur, int k){
        if(cur == NULL)return false;
        return search(root, cur, k - cur->val) || dfs(root, cur->left, k) || dfs(root, cur->right, k);
    }

    bool search(TreeNode* root, TreeNode *cur, int value){
        if(root == NULL)return false;
        return (root->val == value) && (root != cur)
            || (root->val < value) && search(root->right, cur, value)
                || (root->val > value) && search(root->left, cur, value);
    }