LeetCode 173. Binary Search Tree Iterator-优快云博客

本文链接：https://blog.youkuaiyun.com/github_34333284/article/details/51237392

本文介绍如何使用栈实现二叉搜索树（BST）的中序遍历迭代器，通过调用next()函数逐步获取BST中的下一个最小元素，并确保操作在平均O(1)时间内完成，同时使用O(h)的内存，其中h为树的高度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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, whereh is the height of the tree.

I first misunderstood this question. I thought when given a nodes, next() should output the next in-order successor.

This one is in-order traverse, but to print one by one by calling next function.... This is a variation of in-order traversal of binary tree.

class BSTIterator {
private:
    stack<TreeNode*> nodes;
public:
    BSTIterator(TreeNode *root) {
        while(root) {
            nodes.push(root);
            root = root->left;
        }
    }

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

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

Another variation of this question : given a target node in binary tree, get the next node in binary tree in-order order.

// binary tree inorder iterator
struct TreeNode {
  TreeNode* left;
  TreeNode* right;
  TreeNode* parent;
}

// get the nextOne in in-order traversal order.
TreeNode* getNode(TreeNode* current) {
  if(current->right != NULL) {
    return inorderTraversal(current->right);
  } else {
    if(current->parent->left == current) return current->parent;
    else {
      TreeNode* tmp = current;
      TreeNode* next = tmp->parent;
      while(next && next->left != tmp) {
        tmp = next;
        next = next->parent;
      }
      return next;
    }
  }
}