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# 173.Binary-Search-Tree-Iterator

## 题目描述

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.
BSTIterator iterator = new BSTIterator(root);
iterator.next(); // return 3
iterator.next(); // return 7
iterator.hasNext(); // return true
iterator.next(); // return 9
iterator.hasNext(); // return true
iterator.next(); // return 15
iterator.hasNext(); // return true
iterator.next(); // return 20
iterator.hasNext(); // return false
Note:
next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called.

## 代码

### Approach #0 Controlled Recursion

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class BSTIterator {
Stack<TreeNode> stack;
public BSTIterator(TreeNode root) {
// Stack for the recursion simulation
this.stack = new Stack<TreeNode>();
// Remember that the algorithm starts with a call to the helper function
// with the root node as the input
_leftmostInorder(root);
}
private void _leftmostInorder(TreeNode root) {
// For a given node, add all the elements in the leftmost branch of the tree
// under it to the stack.
while (root != null) {
this.stack.push(root);
root = root.left;
}
}
/**
* @return the next smallest number
*/
public int next() {
// Node at the top of the stack is the next smallest element
TreeNode topmostNode = this.stack.pop();
// Need to maintain the invariant. If the node has a right child, call the
// helper function for the right child
if (topmostNode.right != null) {
this._leftmostInorder(topmostNode.right);
}
}
/**
* @return whether we have a next smallest number
*/
public boolean hasNext() {
return this.stack.size() > 0;
}
}

### Approach 1: Using Stack

public class BSTIterator {
private Stack<TreeNode> stack = new Stack<>();
public BSTIterator(TreeNode root) {
while (root != null) {
stack.push(root);
root = root.left;
}
}
public boolean hasNext() {
return !stack.isEmpty();
}
public TreeNode next() {
TreeNode curt = stack.peek();
TreeNode node = curt;
if (node.right == null) {
node = stack.pop(); // 右子树为空才会弹出
while (!stack.isEmpty() && stack.peek().right == node) {
node = stack.pop();
}
} else {
node = node.right;
while (node != null) {
stack.push(node);
node = node.left;
}
}
return curt; // 注意题目其实要求返回的是 curt.val
}
}

### Approach 2:

class BSTIterator {
private Stack<TreeNode> stack = new Stack<>();
TreeNode next = null;
public BSTIterator(TreeNode root) {
next = root;
}
public boolean hasNext() {
while (next != null) {
stack.push(next);
next = next.left;
}
return !stack.isEmpty();
}
while (root != null) {
stack.push(root);
root = root.left;
}
}
public TreeNode next() {
if (!hasNext()) {
return null;
}
TreeNode cur = stack.pop();
next = cur.right;
return cur;
}
}

### Approach #3 Flattening the BST

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class BSTIterator {
ArrayList<Integer> nodesSorted;
int index;
public BSTIterator(TreeNode root) {
// Array containing all the nodes in the sorted order
this.nodesSorted = new ArrayList<Integer>();
// Pointer to the next smallest element in the BST
this.index = -1;
// Call to flatten the input binary search tree
this._inorder(root);
}
private void _inorder(TreeNode root) {
if (root == null) {
return;
}
this._inorder(root.left);
this._inorder(root.right);
}
/**
* @return the next smallest number
*/
public int next() {
return this.nodesSorted.get(++this.index);
}
/**
* @return whether we have a next smallest number
*/
public boolean hasNext() {
return this.index + 1 < this.nodesSorted.size();
}
}