p173

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LeetCode p173 Binary Search Tree Iterator 题解

1.题目：

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.

题意：

给一个平衡二叉树。返回下一个最小的数。

2.解题思路:

见代码，先将数整理成有序的表。

3.代码

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

public class BSTIterator {

    public  TreeNode root;

	public  List<Integer> ans = new ArrayList<Integer>();
	public  int j = 0;

	public void add(TreeNode root) {

		if (root == null)
			return;
		add(root.right);
		j++;
		ans.add(root.val);
		//System.out.println(j+"  "+root.val);
		add(root.left);

	}
    public BSTIterator(TreeNode root) {
        this.root=root;
        add(root);
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return j != 0;
    }

    /** @return the next smallest number */
    public int next() {
        j--;
        System.out.println(j);
		return ans.get(j);
    }
}

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

4.一些总结：