【LeetCode】104. Maximum Depth of Binary Tree 解法及注释,递归,深度搜索

104. Maximum Depth of Binary Tree

Total Accepted: 137961 Total Submissions: 289093 Difficulty: Easy

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.


【分析】

    这个题很简单,就是一个搜索问题,对二叉树进行遍历并记录最大深度即可解决问题,我用的是中序遍历,当然其它遍历也可解决问题。遍历的思路如是,代码的书写风格亦会影响效率,从提高扩展性的角度看,选择一种中序遍历是不错的,作为独立函数执行效率尚可。

【解法及注释】

方法一:递归遍历所有结点,记录最大深度

class Solution {
public:
    int maxDepth(TreeNode* root) 
    {
        int maxdeep=0;
        DFS(root, 0, maxdeep);
        return maxdeep;
    }
    
    void DFS(TreeNode* root, int deepth,int &maxdeep)
    {
        if(root==NULL)return;
        if(deepth+1>maxdeep)maxdeep=deepth+1;
        DFS(root->left,deepth+1,maxdeep);
        DFS(root->right,deepth+1,maxdeep);
    }
};

方法二:网上解法,效率不高

class Solution {
public:
    int maxDepth(TreeNode *root) {
        if (root == NULL) return 0;
        TreeNode* left_node = root->left;
        TreeNode* right_node = root->right;
        int left_dep = maxDepth(left_node);
        int right_dep = maxDepth(right_node);
        return 1 + ((left_dep > right_dep) ? left_dep:right_dep); 
    }
};







你可能感兴趣的:(LeetCode)