[Leetcode] Minimum Depth of Binary Tree
2016-10-04 10:17
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本题目我自己编写了一个代码,但是运行速度很慢,看了别人的代码才知道,有些树的递归逻辑的判断并不需要一定在本次执行,如果函数内部写了异常处理,即对树root为空的处理,则对叶子节点的判断可以由自身的返回值判断,这样就会大大减少代码的体积。
现贴下我自己和他人的代码以供学习
精简版
我用了4个if,而精简版只用了3个,其对叶子节点的判断,巧妙的利用了排除法,减少了一次判断过程,很巧妙~学习学习!!
现贴下我自己和他人的代码以供学习
class Solution { public: int minDepth(TreeNode* root) { if(!root) return 0;//这里已经有了孩子节点为空的判断,因此,叶子节点可以用叶子节点孩子均为空判断 int left= 65536; int right= 65536; if(!root->left&&!root->right) return 1; if(root->left) left = minDepth(root->left); if(root->right) right = minDepth(root->right); return left >= right?right+1:left+1; } };
精简版
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int minDepth(TreeNode *root) { // Start typing your C/C++ solution below // DO NOT write int main() function return minRec(root); } int minRec( TreeNode * root) { if(!root) return 0; int left = minRec( root->left); int right = minRec( root->right); if(left && right) return 1 + min(left, right); if(left || right) return 1+left+right; return 1;//排除了上面两个条件,剩下的只能是叶子节点 } };
我用了4个if,而精简版只用了3个,其对叶子节点的判断,巧妙的利用了排除法,减少了一次判断过程,很巧妙~学习学习!!
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