《剑指offer》刷题笔记(知识迁移能力):二叉树的深度



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。

解题思路:

DFS:题目很简单,用递归遍历只需一行。如果传入的树指针为空指针,则直接返回0;否则,递归调用TreeDepth,遍历左右子树并返回最大值。
BFS:层次遍历,需要使用队列。如果队列不为空,则在循环内不断的pop根节点、push左右子树,同时累加depth。

C++版代码实现:

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/*
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};*/
class Solution {
public:
int TreeDepth(TreeNode* pRoot)
{
return pRoot ? max(TreeDepth(pRoot->left),TreeDepth(pRoot->right))+1 :0;
}
};

层序遍历:

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/*
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};*/
class Solution {
public:
int TreeDepth(TreeNode* pRoot)
{
if(!pRoot) return 0;
queue<TreeNode*> que;
int depth = 0;
que.push(pRoot);
while(!que.empty()){
int size = que.size();
depth++;
for(int i=0; i < size; i++){
TreeNode* node = que.front();
que.pop();
if(node->left) que.push(node->left);
if(node->right) que.push(node->right);
}
}
return depth;
}
};

Python版代码实现:

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# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def TreeDepth(self, pRoot):
# write code here
if not pRoot:
return 0
return max(self.TreeDepth(pRoot.left),self.TreeDepth(pRoot.right))+1

层次遍历:

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# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def TreeDepth(self, pRoot):
# write code here
if not pRoot:
return 0
a=[pRoot]
depth=0
while a:
b=[]
for node in a:
if node.left:
b.append(node.left)
if node.right:
b.append(node.right)
a=b
depth=depth+1
return depth

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完的汪(∪。∪)。。。zzz

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