Code & Fun

第46天。

今天出游,挑到水题Maximum Depth of Binary Tree:

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.

说是水题,就不讲怎么做了,直接上代码吧:

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int maxDepth(TreeNode* root) {
return maxDepth(root,0);
}
int maxDepth(TreeNode *root,int depth) {
if (root == nullptr) return depth;
return max(maxDepth(root->left,depth+1),maxDepth(root->right,depth+1));
}

恩,突然发现好像没必要写的那么长:

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int maxDepth(TreeNode *root) {
if (root == nullptr) return 0;
return max(maxDepth(root->left),maxDepth(root->right))+1;
}

恩,送上一个dicuss中BFS的解法:

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int maxDepth(TreeNode *root)
{
if(root == NULL)
return 0;

int res = 0;
queue<TreeNode *> q;
q.push(root);
while(!q.empty())
{
++ res;
for(int i = 0, n = q.size(); i < n; ++ i)
{
TreeNode *p = q.front();
q.pop();

if(p -> left != NULL)
q.push(p -> left);
if(p -> right != NULL)
q.push(p -> right);
}
}

return res;
}

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