【树】173. 二叉搜索树迭代器

题目：

实现一个二叉搜索树迭代器。你将使用二叉搜索树的根节点初始化迭代器。

调用 next() 将返回二叉搜索树中的下一个最小的数。

 

示例：

 

BSTIterator iterator = new BSTIterator(root);
iterator.next(); // 返回 3
iterator.next(); // 返回 7
iterator.hasNext(); // 返回 true
iterator.next(); // 返回 9
iterator.hasNext(); // 返回 true
iterator.next(); // 返回 15
iterator.hasNext(); // 返回 true
iterator.next(); // 返回 20
iterator.hasNext(); // 返回 false

 

解答：

方法一：中序遍历

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

    List<Integer> list;
    int index;
    public BSTIterator(TreeNode root) {
        this.list = new ArrayList<>();
        this.index = 0;
        inOrder(root);
    }

    public void inOrder(TreeNode root){
        if(root == null) return;
        inOrder(root.left);
        list.add(root.val);
        inOrder(root.right);
    }
    
    /** @return the next smallest number */
    public int next() {
        int res = list.get(index);
        index++;
        return res;
    }
    
    /** @return whether we have a next smallest number */
    public boolean hasNext() {

        return(index<list.size());
    }
}

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator obj = new BSTIterator(root);
 * int param_1 = obj.next();
 * boolean param_2 = obj.hasNext();
 */

 

 

方法二：受控递归（leetcode标答）

算法：

 

 

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

    Stack<TreeNode> stack;

    public BSTIterator(TreeNode root) {
        
        // Stack for the recursion simulation
        this.stack = new Stack<TreeNode>();
        
        // Remember that the algorithm starts with a call to the helper function
        // with the root node as the input
        this._leftmostInorder(root);
    }

    private void _leftmostInorder(TreeNode root) {
      
        // For a given node, add all the elements in the leftmost branch of the tree
        // under it to the stack.
        while (root != null) {
            this.stack.push(root);
            root = root.left;
        }
    }

    /**
     * @return the next smallest number
     */
    public int next() {
        // Node at the top of the stack is the next smallest element
        TreeNode topmostNode = this.stack.pop();

        // Need to maintain the invariant. If the node has a right child, call the 
        // helper function for the right child
        if (topmostNode.right != null) {
            this._leftmostInorder(topmostNode.right);
        }

        return topmostNode.val;
    }

    /**
     * @return whether we have a next smallest number
     */
    public boolean hasNext() {
        return this.stack.size() > 0;
    }
}