剑指Offer 38. 二叉树的深度 （二叉树）

题目描述

输入一棵二叉树，求该树的深度。从根结点到叶结点依次经过的结点（含根、叶结点）形成树的一条路径，最长路径的长度为树的深度。

题目地址

https://www.nowcoder.com/practice/435fb86331474282a3499955f0a41e8b?tpId=13&tqId=11191&rp=2&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking

思路

思路1： 递归，结点为空时，返回0

不为0时，遍历左右子树，在较深的子树上加1

思路2：bfs 层次遍历

每遍历完一层，层数加1

q为当前层的结点，遍历q中的每个结点，然后用tmp保存当前每个结点可能有的左结点和右结点。

Python

# -*- coding:utf-8 -*-
class TreeNode:
    def __init__(self,x):
        self.val = x
        self.left = None
        self.right = None
node1 = TreeNode(8)
node2 = TreeNode(6)
node3 = TreeNode(10)
node4 = TreeNode(5)
node5 = TreeNode(7)
node6 = TreeNode(9)
node1.left = node2
node1.right = node3
node2.left = node4
node2.right = node5
node3.left = node6
class Solution:
    def TreeDepth(self, pRoot):
        # write code here
        # 思路1
        # if not pRoot:
        #     return 0
        # return max(self.TreeDepth(pRoot.left),self.TreeDepth(pRoot.right))+1

        # 思路2
        if not pRoot:
            return 0
        count = self.levelOrder(pRoot)
        return count

    def levelOrder(self,pRoot):
        count = 0
        if not pRoot:
            return count
        queue = []
        queue.append(pRoot)
        while queue:
            temp = []
            for i in range(len(queue)):
                r = queue.pop(0)
                if r.left:
                    queue.append(r.left)
                if r.right:
                    queue.append(r.right)
                temp.append(r.val)
            if temp:
                count += 1
        return count


if __name__ == '__main__':
    result = Solution().TreeDepth(node1)
    print(result)