leetcode 104 Maximum Depth of Binary Tree
阿新 • • 發佈:2018-12-26
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.
題目的意思是求二叉樹的深度,難度的easy。博主在刷的過程中雖然是一遍就AC了,但是程式碼比較醜陋,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); } };
思路就是通過常規的dfs遍歷每個節點,如果ret小於當前節點的深度,則ret變為當前的深度。提交後看了別人的解法,深覺程式碼簡潔優雅,結題思路一致,搬運如下:
int maxDepth(TreeNode *root) {
if (root == null) {
return 0;
}
return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
}