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.
题解：最开始有点误解，其实就是按照从小到大iterator这个树。类似于中序遍历
每次弹出一个node，则将node的right和right的所有左branch压栈
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class BSTIterator {
private Stack<TreeNode> stack = new Stack<>();
public BSTIterator(TreeNode root) {
pushAll(root);
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !stack.isEmpty();
}
/** @return the next smallest number */
public int next() {
TreeNode tmpNode = stack.pop();
pushAll(tmpNode.right);
return tmpNode.val;
}
private void pushAll(TreeNode root){
for(; root!=null; stack.push(root), root = root.left);
}
}
/**
* Your BSTIterator will be called like this:
* BSTIterator i = new BSTIterator(root);
* while (i.hasNext()) v[f()] = i.next();
*/
二刷
思路同上
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class BSTIterator {
private Stack<TreeNode> stack;
public BSTIterator(TreeNode root) {
stack = new Stack<>();
pushLeft(root);
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !stack.isEmpty();
}
/** @return the next smallest number */
public int next() {
TreeNode cur = stack.pop();
pushLeft(cur.right);
return cur.val;
}
private void pushLeft(TreeNode root){
while(root!=null){
stack.push(root);
root = root.left;
}
}
}
/**
* Your BSTIterator will be called like this:
* BSTIterator i = new BSTIterator(root);
* while (i.hasNext()) v[f()] = i.next();
*/
