173. Binary Search Tree Iterator

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling next() will return the next smallest number in the BST.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

S: BST in order traversal

1. create empty stack "nodes"
2. initiate current node "node" as root;
3. push current node to stack and set cur node to node->left until node is empty
4. if cur node is not empty, go to 3, else: cur node = stack.top(), stack.pop(), print cur node value and set node = node->right
5. if stack.emty and node = NULL, we are done.

time complexity of next() is O(n) but amortized O(1) time, since next got called n times at most, each node at most get visited twice. space complexity O(h）h is the depth of the tree

class BSTIterator {
public:
    BSTIterator(TreeNode *root) {
        node = root;
    }

    /** @return whether we have a next smallest number */
    bool hasNext() {
        return (!nodes.empty() || node);
    }

    /** @return the next smallest number */
    int next() {
        TreeNode *cur;
        while(node) {
            nodes.push(node);
            node = node->left;
        }
        node = nodes.top();
        nodes.pop();
        cur = node;
        node = node->right;
        return cur->val;
    }
private:
    TreeNode *node;
    stack<TreeNode*> nodes;
};

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */

501. Find Mode in Binary Search Tree

Given a binary search tree (BST) with duplicates, find all themode(s)(the most frequently occurred element) in the given BST.

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than or equal to the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

For example:
Given BST[1,null,2,2],

   1
    \
     2
    /
   2

return[2].

Note:If a tree has more than one mode, you can return them in any order.

Follow up:Could you do that without using any extra space? (Assume that the implicit stack space incurred due to recursion does not count).

S: inorder traversal O(n)

中序遍历，计算最大频率。因为需要O(1)space，第二次中序遍历找到结果

class Solution {
public:
    vector<int> findMode(TreeNode* root) {
        int mode = 0, count = 0;
        findModeDfs(root, nullptr, mode, count);
        vector<int> result;
        count = 0;
        getMode(root, nullptr, mode, count, result);
        return result;
    }

    TreeNode* findModeDfs(TreeNode* root, TreeNode* pre, int& mode, int& count) {
        if (!root) return nullptr;
        if (root->left) pre = findModeDfs(root->left, pre, mode, count);
        count = (pre && pre->val == root->val)? count + 1 : 1;
        mode = max(mode, count);
        return root->right? findModeDfs(root->right,root, mode, count) : root;
    }

    TreeNode* getMode(TreeNode* root, TreeNode* pre, int mode, int& count, vector<int>& result) {
        if (!root) return nullptr;
        if (root->left) pre = getMode(root->left, pre, mode, count, result);
        count = pre && pre->val == root->val? count + 1 : 1;
        if (mode == count) result.push_back(root->val);
        return root->right? getMode(root->right, root, mode, count, result) : root;
    }
};