递归与二叉树_leetcode222

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

class Solution1(object):
    def countNodes(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0

        return self.countNodes(root.left)+self.countNodes(root.right)+1

# https: // segmentfault.com / a / 1190000011040217
# https://blog.csdn.net/qq_17550379/article/details/82052746

# 如果右子树的高度等于整棵二叉树的高度-1的话，那么左子树一定是一棵满二叉树，
# 这个时候我们就很容易的计算出总结点数nodes=2**(h)-1 + 1 +右子树节点数（这里的+1表示root节点）。
# 如果右子树的高度不等于整棵二叉树的高度-1的话，那么左子树不一定是一棵满二叉树，但是有一点可以确定，
# 右子树一定是一棵满二叉树，这个时候我们就很容易的计算出总结点数nodes=2**(h-1)-1 + 1 +左子树节点数（这里的+1表示root节点）。
# 根据这个思路我们只要不断循环下去直到root==None结束。




class Solution2(object):
    def countNodes(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        def height(t):
            h = -1
            while t:
                h += 1
                t = t.left
            return t

        h = height(root)

        nodes = 0

        while root:

             if height(root.right) == h-1:
                 nodes += 2**h
                 root = root.right
             else :
                 nodes += 2**(h-1)
                 root = root.left

             h -= 1

        return nodes



class Solution3(object):
    def countNodes(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """

        def getHeight(node):
            height = 0
            tempNode = node

            while tempNode:
                height += 1
                tempNode = tempNode.left
            return height

        nodes = 0

        tempRoot = root

        while tempRoot:

            height = getHeight(tempRoot)

            rightHeight = getHeight(tempRoot.right)


            if rightHeight == height-1:
                nodes += 2 ** (height-1)
                tempRoot = tempRoot.right

            else:
                nodes += 2 ** (height-2)
                tempRoot = tempRoot.left

        return nodes