173. Binary Search Tree Iterator(中序遍历)

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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.

 

 

class BSTIterator {
public:
    stack<TreeNode*> stk;
    TreeNode* cur;
    BSTIterator(TreeNode* root) {
        cur = root;
    }
    
    int next() {
        int ret;
        while(cur != nullptr) {
            stk.push(cur);
            cur = cur->left;
        }
        cur = stk.top();
        stk.pop();
        ret = cur->val;
        cur = cur->right;
        return ret;
    }
    
    bool hasNext() {
        return (cur != nullptr || !stk.empty());
    }
};

 

 

 

 

 

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

public class BSTIterator {
    private Stack<TreeNode> stack = new Stack<TreeNode>();
    public BSTIterator(TreeNode root) {
        pushAll(root);
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.isEmpty();
    }

    /** @return the next smallest number */
    public int next() {
        TreeNode node = stack.pop();
        pushAll(node.right);
        return node.val;
        
    }
    private void pushAll(TreeNode node){
        while(node != null){
            stack.push(node);
            node = node.left;
        }
            
    }
}

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */