【python/leetcode/M】Binary Search Tree Iterator

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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.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.

实现思路

要找到二叉搜索树中的最小节点，应该从根节点递归遍历左节点，直到遍历的节点没有左节点，那么该节点就是二叉树中的最小节点。现在已经有二叉搜索树中没有访问过的最小节点了，那么当访问了该节点后，剩余没有访问的树中最小的节点在哪里呢？如果该节点有右子树，那么在它的右子树中（又回到了找一棵二叉搜索树的最小元素，不过这棵二叉搜索树变小了）；如果没有右子树，那么就是它的父节点。

为了能够快速定位到父节点，我们可以用栈将遍历路径暂存起来，当进行next()操作时，我们弹出栈顶元素并进行访问，如果它有右子树的话就遍历它的右子树；如果没有右子树，当下次出栈操作时就是访问当前节点的父节点了。

实现代码

# Definition for a  binary tree node
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class BSTIterator(object):
    def __init__(self, root):
        """
        :type root: TreeNode
        """
        self.stack = []
        self._pushLeft(root)
        

    def hasNext(self):
        """
        :rtype: bool
        """
        return self.stack
        

    def next(self):
        """
        :rtype: int
        """
        node = self.stack.pop()
        self._pushLeft(node.right)
        return node.val
    
    
    def _pushLeft(self,node):
        while node:
            self.stack.append(node)
            node = node.left
            
# Your BSTIterator will be called like this:
# i, v = BSTIterator(root), []
# while i.hasNext(): v.append(i.next())