# Trees topic

# 98. Validate Binary Search Tree
## Question: (Medium)
Given the root of a binary tree, determine if it is a valid binary search tree (BST).

A valid BST is defined as follows:

* The left subtree of a node contains only nodes with keys less than the node's key.
* The right subtree of a node contains only nodes with keys greater than the node's key.
* Both the left and right subtrees must also be binary search trees.
## Solution: (C++)

// 利用inorder traversal來解
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> record;
    void inorderTraversal(TreeNode* root){
        if(root==NULL)
            return; 
        inorderTraversal(root->left);
        record.push_back(root->val);
        inorderTraversal(root->right);
    }
    bool isValidBST(TreeNode* root) {
        inorderTraversal(root);
        for(int i=1; i<record.size(); i++){
            if(record[i]<=record[i-1])
                return false;
        }
        return true;
    }
};

// using recursive 
class Solution {
public:
    bool check_node(TreeNode* root, long long int left, long long int right){
        if(root==NULL)
            return true;
        if(root->val>left && root->val<right){
            return check_node(root->right, root->val, right) && check_node(root->left, left, root->val);
        }
        else{
            return false;
        }
    }
    bool isValidBST(TreeNode* root) {
        long long int maxima = 12147483648;
        long long int minima = -12147483648;
        cout<<minima<<" "<<maxima<<endl;
        if(root->left == NULL && root->right == NULL)
            return true;
        else
            return check_node(root, minima, maxima);
    }
};

## Notes:
1. 個人覺得用inorder來解比較直觀，因為inorder的output必定是由小到大，所以只需要檢查順序即可
2. recursive則是寫一個function，判斷該node是否符合條件，有沒有大於left，有沒有小於right
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# 2265. Count Nodes Equal to Average of Subtree
## Question (Medium)
Given the root of a binary tree, return the number of nodes where the value of the node is equal to the average of the values in its subtree.

Note:

The average of n elements is the sum of the n elements divided by n and rounded down to the nearest integer.
A subtree of root is a tree consisting of root and all of its descendants.

## Solution (C++)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int ans = 0;
    void post_order(TreeNode* node, int& sum, int& count){
        if(node == NULL){
            return ;
        }
        post_order(node->left, sum, count);
        int temp = count;
        int temp_sum = sum;
        sum-=temp_sum; count-=temp;
        post_order(node->right, sum, count);
        count+=temp; sum+=temp_sum;
        count++; sum+=node->val;
        if(node->val == floor(double(sum/count))){
            ans++;
        }
    }
    int averageOfSubtree(TreeNode* root) {
        int sum = 0, count = 0;
        post_order(root, sum, count);
        return ans;
    }
};

## Notes:
* 這題用postorder先算出left subtree的總和和次數，再計算right subtree的總和和次數
* 比較麻煩的點是在遍歷right subtree不需要考慮left subtree的總和和node數量。
* 所以需要做一點改變(line 20~24)，我們在遍歷完left subtree時，先把summation的值減掉，遍歷完right subtree的時候，再把它加回來。
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## 103. Binary Tree Zigzag Level Order Traversal

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        queue<pair<TreeNode*, int>> q;
        vector<vector<int>> ans;
        q.push(make_pair(root, 0));
        while(q.size()!=0){
            int qs = q.size();
            vector<int> temp_arr;
            for(int i=0; i<qs; i++){
                pair<TreeNode*, int> temp = q.front(); q.pop();
                if(temp.first == NULL)
                    continue;
                if(temp.second%2==0)
                    temp_arr.push_back(temp.first->val);
                else
                    temp_arr.insert(temp_arr.begin(), temp.first->val);
                q.push(make_pair(temp.first->left, temp.second+1));
                q.push(make_pair(temp.first->right, temp.second+1));
                
            }
            if(temp_arr.size()>0)
                ans.push_back(temp_arr);
        }
        return ans;
    }
};

### Notes:
1. level order traversal
2. 其實就是bfs，只是要注意level，如果是奇數層則要反過來traverse。