LeetCode-173. Binary Search Tree Iterator (JAVA)BST的迭代

173. 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, whereh is the height of the tree.

用一个Stack记录从根节点到当前节点的路径。next的时候就返回Stack最上面的元素。不过拿出最上面的元素后，我们还要看一下这个被返回的元素是否有右节点，如果有的话，就把它的右节点及右节点的所有左边节点都压入栈中。另外，初始化栈时，我们要找到最左边的节点，也就是中序遍历的第一个节点，并把根到第一个节点的路径都压入栈。

本题目的本质是二叉树的中序遍历

public class BSTIterator {private Stack<TreeNode> stk;public BSTIterator(TreeNode root) {stk = new Stack<>();// 先找到第一个节点，并把路径入栈while (root != null) {stk.push(root);root = root.left;}}/** * @return whether we have a next smallest number */public boolean hasNext() {// 栈为空时不再有下一个return !stk.isEmpty();}/** * @return the next smallest number */public int next() {TreeNode curr = stk.pop();int res = curr.val;// 如果该元素有右节点，把它的右节点及右节点的所有左边节点都压入栈中curr = curr.right;while (curr != null) {stk.push(curr);curr = curr.left;}return res;}}

二叉树的中序遍历非递归

public List<Integer> inorderTraversal(TreeNode root) {          List<Integer> list = new ArrayList<>();          if (root == null)              return list;          Stack<TreeNode> stack = new Stack<>();          while (root != null || !stack.isEmpty()) {              if (root != null) {                  stack.push(root);                  root = root.left;              } else {                  root = stack.pop();                  list.add(root.val);                  root = root.right;              }          }          return list;      }