如何递归地找到二叉树中节点的高度 [英] how to find the height of a node in binary tree recursively

问题描述

path = 0 # the lenght of the path
    while self.right != None or self.left != None:
        while self.right != None:
            self = self.right
            path = path +1 
        while self.left != None:
            self = self.left
            path = path +1
    return path

这是我用于查找高度的示例代码，定义为 从自身到叶子的节点数的最长路径.叶节点的高度为1.

this is my sample code for find the Height, is defined as the length of the longest path by number of nodes from self to a leaf. The height of a leaf node is 1.

它不起作用.

推荐答案

您正在做的事情不是递归的，而是迭代的. 递归类似于:

What you're doing isn't recursive, it's iterative. Recursive would be something like:

def height(node):
    if node is None:
        return 0
    else:
        return max(height(node.left), height(node.right)) + 1

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