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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.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
iterative inorder traveral 的变型。
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 // iterative inorder traveral change 12 public class BSTIterator { 13 LinkedList<TreeNode> stack; 14 boolean flg; 15 TreeNode tmp; 16 public BSTIterator(TreeNode root) { 17 stack = new LinkedList<TreeNode> (); 18 tmp = root; 19 flg = (root != null); 20 } 21 22 /** @return whether we have a next smallest number */ 23 public boolean hasNext() { 24 return flg && !(stack.isEmpty() && tmp == null); 25 } 26 27 /** @return the next smallest number */ 28 public int next() { 29 while(true){ 30 if(tmp != null){ 31 stack.push(tmp); 32 tmp = tmp.left; 33 }else{ 34 TreeNode result = stack.pop(); 35 tmp = result.right; 36 return result.val; 37 } 38 } 39 } 40 } 41 42 /** 43 * Your BSTIterator will be called like this: 44 * BSTIterator i = new BSTIterator(root); 45 * while (i.hasNext()) v[f()] = i.next(); 46 */
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原文地址:http://www.cnblogs.com/reynold-lei/p/4332691.html