BST迭代器

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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.

先实现中序遍历代码。然后改造为迭代器。

public class BSTIterator {
        private TreeNode node;
        private Stack<TreeNode> stack;

        public BSTIterator(TreeNode root) {
            node = root;
            stack = new Stack<TreeNode>();
            while (node != null) {
                stack.push(node);
                node = node.left;
            }
        }

        /**
         * @return whether we have a next smallest number
         */
        public boolean hasNext() {
            return !stack.empty();

        }

        /**
         * @return the next smallest number
         */
        public int next() {

            TreeNode t = node = stack.pop();
            if (node.right != null) {
                node = node.right;
                while (node != null) {
                    stack.push(node);
                    node = node.left;
                }
            }
            return t.val;
        }

        public void inOrder(TreeNode root) {
            TreeNode node = root;

            while (node != null) {
                stack.push(node);
                node = node.left;
            }
            while (!stack.empty()) {
                System.out.print((node = stack.pop()) + ",");
                if (node.right != null) {
                    node = node.right;
                    while (node != null) {
                        stack.push(node);
                        node = node.left;
                    }
                }
            }

        }
    }

BST迭代器

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原文地址：http://blog.csdn.net/wongson/article/details/45076829