标签：and 二叉树的遍历 时间 div run 时间复杂度 判断 blog 遍历
题目：
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.
题意及分析：给出一个二叉排序树，求一个该二叉树的遍历器，满足：(1)hasNext()判断树中是否存在一个除了当前数的最小数。 (2)next()返回树中除了当前数的最小数。(3)要求o(1)的时间复杂度和o(h)的空间复杂度。使用一个栈保存即可，初始保存从根节点到最左节点的值，对于next，如果stack非空就存在hasnext smallest number；对于next，下一个最小的就是栈顶的值（因为该栈顶点的左子节点就是当前点，而栈顶点的右子树上的点肯定比栈顶点大），取出栈顶的点，然后对该点的右子节点左上面的操作(即遍历该点到该点的最左叶节点)。
代码：
/**
* 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<>();
TreeNode cur = root;
while(cur != null){
stack.push(cur);
if(cur.left != null)
cur = cur.left;
else
break;
}
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !stack.isEmpty();
}
/** @return the next smallest number */
public int next() {
TreeNode node = stack.pop();
TreeNode cur = node;
// traversal right branch
if(cur.right != null){
cur = cur.right;
while(cur != null){
stack.push(cur);
if(cur.left != null)
cur = cur.left;
else
break;
}
}
return node.val;
}
}
/**
* Your BSTIterator will be called like this:
* BSTIterator i = new BSTIterator(root);
* while (i.hasNext()) v[f()] = i.next();
*/
[LeetCode] 173. Binary Search Tree Iterator Java
标签：and 二叉树的遍历 时间 div run 时间复杂度 判断 blog 遍历
原文地址：http://www.cnblogs.com/271934Liao/p/7029115.html
