OJ练习24——T104 Maximum Depth of Binary Tree

求二叉树深度。

【思路】

很简单，二叉树经典。

用递归求左树右树的深度，较大值加1即可。

【other code】

int maxDepth(TreeNode *root) {
        if(root==NULL)
            return 0;
        //if(maxDepth(root->left)>=maxDepth(root->right))
            //return maxDepth(root->left)+1;
        //else
            //return maxDepth(root->right)+1;
        int max1=maxDepth(root->left);
        int max2=maxDepth(root->right);
        return max1>max2?max1+1:max2+1;
        //return max(maxDepth(root->left), maxDepth(root->right))+1;
        //return maxDepth(root->left)>maxDepth(root->right)?maxDepth(root->left)+1:maxDepth(root->right)+1;
    }

【问题】

注释的是我原来的写法，发现要么用int max1 max2分别表示后，再用条件选择return

要么直接用max判断递归调用的最大值，否则就会报“超时”错，一长串。

哦！！！我晓得了！！！

直接用条件选择如：return maxDepth(root->left)>maxDepth(root->right)?maxDepth(root->left)+1:maxDepth(root->right)+1;

其实在判断maxDepth(root->left)>maxDepth(root->right)和后面计算maxDepth(root->left)+1是进行了两次递归！！当然是没有返回的！

原来如此，代码果真不诳我！

时间： 2024-08-02 13:23:40

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