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.
思路:
代码:
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 11 public class BSTIterator { 12 13 private TreeNode cur; 14 private Stack<TreeNode> stack; 15 16 public BSTIterator(TreeNode root) { 17 cur = root; 18 stack = new Stack<>(); 19 20 } 21 22 /** @return whether we have a next smallest number */ 23 public boolean hasNext() { 24 if(!stack.isEmpty()|| cur!=null) return true; 25 return false; 26 } 27 28 /** @return the next smallest number */ 29 public int next() { 30 while(cur!=null){ 31 stack.push(cur); 32 cur = cur.left; 33 } 34 cur = stack.pop(); 35 int val = cur.val; 36 cur = cur.right; 37 return val; 38 } 39 } 40 41 /** 42 * Your BSTIterator will be called like this: 43 * BSTIterator i = new BSTIterator(root); 44 * while (i.hasNext()) v[f()] = i.next(); 45 */