LeetCode 之 Maximum Depth of Binary Tree
2013-11-05 14:22
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最近看了点动态规划,二叉树的深度用递归方法解决,算得上是动态规划,不过由于二叉树比较简单,因此不需要应用带备份的算法,而且由于二叉树有方向,只能自顶向下,不能自底向上,说是动态规划,其实就是递归而已。。。。不过学了东西之后归类还是好的,而且在开始犯了一个2b的错误,为了简化代码下了如下错误的:
当然,二叉树的深度计算有多种方法,其实最容易想到的应该是压栈,这个就不用递归,不过代码应该比较复杂,由于c++常年不看,压栈都不太会了,实在不行可以用链表代替栈(又一个2B的想法),不过还是用递归是最简洁的代码,不过本人水平有限,别的简便的方法还有没有就不清楚了。。。。
正确的:
class Solution { public: int maxDepth(TreeNode *root) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. return depth(root); } int depth (TreeNode*root){ if(root == NULL){ return 0; } //2b的开始,导致depth左节点和右节点都计算了2次,直接导致超时,其实记下来就ok啦! return ( depth(root->left) >depth(root->right)? depth(root->left) +1:depth(root->right)+1); } };
当然,二叉树的深度计算有多种方法,其实最容易想到的应该是压栈,这个就不用递归,不过代码应该比较复杂,由于c++常年不看,压栈都不太会了,实在不行可以用链表代替栈(又一个2B的想法),不过还是用递归是最简洁的代码,不过本人水平有限,别的简便的方法还有没有就不清楚了。。。。
正确的:
class Solution { public: int maxDepth(TreeNode *root) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. return depth(root); } int depth (TreeNode*root){ if(root == NULL){ return 0; } int left = depth(root->left) ; int right = depth(root->right); return (left>right?left+1:right+1); } };
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