LeetCode104. Maximum Depth of Binary Tree(二叉树 / 搜索)
- Easy
- Accepted:458,799
- Submissions:774,287
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.
Note: A leaf is a node with no children.
Example:
Given binary tree [3,9,20,null,null,15,7]
,
3
/ \
9 20
/ \
15 7
return its depth = 3.
链接
https://leetcode.com/problems/maximum-depth-of-binary
题意
给定一棵二叉树,求它的最大深度。即从根节点到叶子节点的最长路径长度。
题解
递归的搜索左子树的深度l1,右子树的深度l2,那么对于当前结点的最大深度就是max(l1, l2) + 1。时间复杂度的话考虑一颗满二叉树,$T(n) = 2T(n/2)+ 1, T(1) = 1$,为$O(n)$。
代码
- Runtime: 16 ms, faster than 99.26% of C++ online submissions for Maximum Depth of Binary Tree.
- Memory Usage: 19.6 MB, less than 12.18% of C++ online submissions for Maximum Depth of Binary Tree.
/**
* 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 maxDepth(TreeNode* root) {
if (root == NULL)
return 0;
int leftMaxDepth = maxDepth(root->left);
int rightMaxDepth = maxDepth(root->right);
return max(leftMaxDepth, rightMaxDepth) + 1;
}
};
// simple
class Solution {
public:
int maxDepth(TreeNode* root) {
return root == NULL ? 0 : max(maxDepth(root->left), maxDepth(root->right)) + 1;
}
};
- 当然我们也可以使用队列,即bfs。
class Solution {
public:
int maxDepth(TreeNode* root) {
if (root == NULL)
return 0;
int ans = 0;
queue <TreeNode*> q;
q.push(root);
while (!q.empty())
{
++ans;
int n = q.size();
for (int i = 0; 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 ans;
}
};
参考
https://leetcode.com/problems/maximum-depth-of-binary-tree/discuss/34207
The end.
2019年3月7日 星期四