找LeetCode上简单的题目来撑过30天啦(DAY9)

题号：104 标题：Maximum Depth of Binary Tree难度：Easy

Given the root of a binary tree, return its maximum depth.

A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Input: root = [3,9,20,null,null,15,7]
Output: 3

Example 2:
Input: root = [1,null,2]
Output: 2

Example 3:
Input: root = []
Output: 0

Example 4:
Input: root = [0]
Output: 1

Constraints:
The number of nodes in the tree is in the range [0, 104].
-100 <= Node.val <= 100

Time Limit Exceeded的程序码

int check(struct TreeNode* root2,int D2){
    if(root2 == NULL){
        return 0;
    }else if(root2->left != NULL || root2 -> right != NULL){
        D2= D2+1;    
        if(check(root2->left,D2) > check(root2->right,D2)){
            D2 = check(root2->left,D2);
        }else{
            D2 = check(root2->right,D2);
        }           
     }
    return D2;
}

int maxDepth(struct TreeNode* root){
    int D = 1;   
    return check(root,D);
}

最终的程序码

int check(struct TreeNode* root2,int D2){
    //printf("%d\n",root2->val);
    if(root2 == NULL){
        return 0;
    }else if(root2->left != NULL || root2 -> right != NULL){
        D2= D2+1;
        int l = check(root2->left,D2);
        int r  =  check(root2->right,D2);
        if(l > r){
            D2 =l;
        }else{
            D2 =r;
        }           
     }
    return D2;
}

int maxDepth(struct TreeNode* root){
    int D = 1;   
    return check(root,D);
}

花比较多的时间
今天花比较多时间在解决Time Limit Exceeded的问题，最後发现我在if里面recursive，导致不必要的call function，改为用变数存return的值再去做判断，就没有这个问题啦

DAY9心得
我想离职了，有没有公司缺人啊