【LeetCode】104. Maximum depth of binary tree

题目

给定一个二叉树,找出其最大深度.
二叉树的深度为根节点到最远叶子节点的最长路径上的节点数.
说明: 叶子节点是指没有子节点的节点.

示例：

给定二叉树 [3,9,20,null,null,15,7],
返回它的最大深度 3 .

题解

使用递归

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */
class Solution {
    
public:
    int fun(TreeNode* node,int high)
    {
    
        if(node==nullptr)
            return high;
        high++;
        return max(fun(node->left,high),fun(node->right,high));
    }


    int maxDepth(TreeNode* root) {
    
        int high=0;
        if(root)
            high = fun(root,high);
        return high;
    }
};


//改进
class Solution {
    
public:
    int maxDepth(TreeNode* root) {
    
        if(root==nullptr)
            return 0;
        return max(maxDepth(root->left),maxDepth(root->right))+1;
    }
};

使用广度优先遍历

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */
class Solution {
    
public:
    int maxDepth(TreeNode* root) {
    
        if(root==nullptr)
            return 0;
        int high = 0;
        queue<TreeNode*> myqueue;
        myqueue.emplace(root);
        while(!myqueue.empty())
        {
    
            high++;
            queue<TreeNode*> temp;
            int len = myqueue.size();
            while(len>0)
            {
    
                len--;
                TreeNode* node = myqueue.front();
                myqueue.pop();
                if(node->left)
                    myqueue.emplace(node->left);
                if(node->right)
                    myqueue.emplace(node->right);
            }
        }
        return high;
    }
};