leetcode Binary Search Tree Iterator

素颜马尾好姑娘i 2022-08-04 16:58 166阅读 0赞

# 题目 #

原题链接：[https://leetcode.com/problems/binary-search-tree-iterator/][https_leetcode.com_problems_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.

# 分析 #

题目不好理解，说白了就是使用非递归的形式中序遍历二叉树，这也是c++的STL中关联容器迭代器的实现原型。非递归中序遍历二叉树的话就是借助栈，让下次输出的目标节点始终存放在栈顶位置。每次输出一个节点，就将此节点的后续节点放入栈中（沿着右子树的左子树一直向左就是下一个要输出的节点）。

# 代码 #

/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */
    class BSTIterator {
    private:
        stack<TreeNode* > help_stack;
    public:
        BSTIterator(TreeNode *root) {
            TreeNode* p = root;
            while(p){
                help_stack.push(p);
                p = p->left;
            }
        }
    
        /** @return whether we have a next smallest number */
        bool hasNext() {
            return !help_stack.empty();
        }
    
        /** @return the next smallest number */
        int next() {
            TreeNode *result = help_stack.top();
            help_stack.pop();
            TreeNode *p = result->right;
            while(p){
                help_stack.push(p);
                p = p->left;
            }
            return result->val;
        }
    };
    
    /** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */

[https_leetcode.com_problems_binary-search-tree-iterator]: https://leetcode.com/problems/binary-search-tree-iterator/