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();
  */