LeetCode: Binary Search Tree Iterator 解题报告

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.

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SOLUTION 1:

使用inorder traversal把tree转化为arraylist.

 递归

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

public class BSTIterator {
    ArrayList<TreeNode> list;
    int index;

    public BSTIterator(TreeNode root) {
        list = new ArrayList<TreeNode>();
        iterator(root, list);
        
        index = 0;
    }
    
    // solution 1: recursion.
    public void dfs (TreeNode root, ArrayList<TreeNode> ret) {
        if (root == null) {
            return;
        }
        
        //Use inorder traversal.
        dfs(root.left, ret);
        ret.add(root);
        dfs(root.right, ret);
    }
    
   
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        if (index < list.size()) {
            return true;
        }
        
        return false;
    }

    /** @return the next smallest number */
    public int next() {
        return list.get(index++).val;
    }
}

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

SOLUTION 2:

the iterator version.

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

public class BSTIterator {
    ArrayList<TreeNode> list;
    int index;

    public BSTIterator(TreeNode root) {
        list = new ArrayList<TreeNode>();
        iterator(root, list);
        
        index = 0;
    }
    
      
    // solution 2: Iterator.
    public void iterator (TreeNode root, ArrayList<TreeNode> ret) {
        if (root == null) {
            return;
        }
        
        Stack<TreeNode> s = new Stack<TreeNode>();
        // bug 1: use push instead of put
        TreeNode cur = root;
                
        while (true) {
            // bug 2: should push the node into the stack.
            while (cur != null) {
                s.push(cur);
                cur = cur.left;
            }
            
            if (s.isEmpty()) {
                break;
            }
            
            // bug 3: should pop a node from the stack.
            // deal with the top node in the satck.            
            cur = s.pop();
            
            // bug 2: should be cur not root.
            ret.add(cur);
            cur = cur.right;
        }
    }
    
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        if (index < list.size()) {
            return true;
        }
        
        return false;
    }

    /** @return the next smallest number */
    public int next() {
        return list.get(index++).val;
    }
}

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