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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

 

这道题主要就是考二叉树的中序遍历的非递归形式,需要额外定义一个栈来辅助,二叉搜索树的建树规则就是左<根<右,用中序遍历即可从小到大取出所有节点。代码如下:

 

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
public:
    BSTIterator(TreeNode *root) {
        while (root) {
            s.push(root);
            root = root->left;
        }
    }

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

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

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

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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