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.
题目的意思是求二叉树的深度,难度的easy。博主在刷的过程中虽然是一遍就AC了,但是代码比较丑陋,如下:
/**
* 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) {
int ret = 0;
if(!root)
return ret;
dfsDepth(ret, 1, root);
return ret;
}
void dfsDepth(int &ret, int curr_depth, TreeNode *root)
{
if(!root)
return;
if(ret < curr_depth)
ret = curr_depth;
dfsDepth(ret, curr_depth + 1, root->left);
dfsDepth(ret, curr_depth + 1, root->right);
}
};
思路就是通过常规的dfs遍历每个节点,如果ret小于当前节点的深度,则ret变为当前的深度。提交后看了别人的解法,深觉代码简洁优雅,结题思路一致,搬运如下:
int maxDepth(TreeNode *root) {
if (root == null) {
return 0;
}
return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
}