
一、前言
本系列文章为《剑指Offer》刷题笔记。
刷题平台:牛客网
书籍下载:共享资源
二、题目
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。
1、思路
这道题蛮简单的,求二叉树的深度。可以是递归的方法,属于DFS(深度优先搜索);另一种方法是按照层次遍历,属于BFS(广度优先搜索)。
2、代码
C++:
DFS方法:
/*
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 == NULL){
return 0;
}
int left = TreeDepth(pRoot->left);
int right = TreeDepth(pRoot->right);
return (left > right) ? (left + 1) : (right + 1);
}
};
BFS方法:
/*
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 == NULL){
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;
}
};
感谢@小小毛提供的本地测试用例:
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
def TreeDepth(self, root):
# write code here
if root is None:
return 0
left=self.TreeDepth(root.left)
right=self.TreeDepth(root.right)
print(left,right)
return max(left,right)+1
if __name__=='__main__':
A1 = TreeNode(1)
A2 = TreeNode(2)
A3 = TreeNode(3)
A4 = TreeNode(4)
A5 = TreeNode(5)
A6 = TreeNode(6)
A1.left=A2
A1.right=A3
A2.left=A4
A2.right=A5
A4.left=A6
solution=Solution()
ans=solution.TreeDepth(A1)
print('ans=',ans)
来源:
https://cuijiahua.com/blog/2018/01/basis_38.html
