Maximum Depth of Binary Tree

本题收获

1.树时使用递归

2.注意边界条件时输出的值,仔细阅读题意,若是面试时,问清边界条件。

  题目:  

  Given a binary tree, find its maximum depth.

  The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

  题目中说:最短路径的定义是从根节点到最近子节点的最短路径,必须是子节点[1,2]正确输出应该是2,如果不判断左子节点和右子节点输出就会变成1.

  [1,2]:       [1,2,3]   

    1        1
   /        /
   2        2
           /
          3

  思路:

    我的思路:首先判断root是否等于NULL,然后直接使用递归。

    leetcode:1.判断root是否为NULL,2.判断root->left是否为空,3.判断root->right是否为空,4.利用递归,每次加1计算长度。

  代码:

 class Solution {
public:
int minDepth(TreeNode* root) {
if (root == NULL) return ;
if (root->left == NULL) return minDepth(root->right) + ;
if (root->right == NULL) return minDepth(root->left) + ;
return min(minDepth(root->left), minDepth(root->right)) + ;
}
};

  出错代码1:

    没有判断左子节点和右子节点为空的情况,若不判断在1,2这种情况就返回1,但是正确应该是返回2,题目是说从根节点到子节点。

 class Solution {
public:
int minDepth(TreeNode* root) {
if (root == NULL) return ;
return min(minDepth(root->left), minDepth(root->right)) + ;
}
};

  我的测试代码:

 // Minimum Depth of Binary Tree.cpp : 定义控制台应用程序的入口点。
// #include "stdafx.h"
#include "iostream"
#include "malloc.h"
#include "algorithm"
using namespace std; struct TreeNode
{
int val;
TreeNode *left, *right;
TreeNode(int x) :val(x), left(NULL), right(NULL){};
}; void creatTree(TreeNode* &T)
{
int data;
cin >> data;
if (data == -)
{
T = NULL;
}
else
{
T = (TreeNode*)malloc(sizeof(TreeNode));
T->val = data;
creatTree(T->left);
creatTree(T->right);
} } class MyClass
{
public:
int minDepth(TreeNode* root)
{ if (root == NULL) return ;
if (root->left == NULL) return minDepth(root->right) + ; //leetcode上1,2这样最短是返回2,而不是1
if (root->right == NULL) return minDepth(root->left) + ; //题目中有说是从根节点到最近的叶子节点,必须时到叶子节点啊
return min(minDepth(root->left), minDepth(root->right)) + ; } }; int _tmain(int argc, _TCHAR* argv[])
{
TreeNode* root = NULL;
int m = ;
creatTree(root);
MyClass solution;
m = solution.minDepth(root);
cout << m << endl;
system("pause");
return ;
}

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