Leetcode—104.二叉树的最大深度【简单】

2023每日刷题（六）

Leetcode—104.二叉树的最大深度

递归实现代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */


int maxDepth(struct TreeNode* root){
    if(!root) {
        return 0;
    }
    int left = maxDepth(root->left);
    int right = maxDepth(root->right);
    return (left > right ? left + 1: right+1);
}
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运行结果

BFS队列实现代码

/**
 * 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 == NULL) {
            return 0;
        }
        int ans = 0;
        queue<TreeNode*> qu;
        qu.push(root);
        while(!qu.empty()) {
            ans++;
            int n = qu.size();
            for(int i = 0; i < n; i++) {
                TreeNode* p = qu.front();
                qu.pop();
                if(p->left != NULL) {
                    qu.push(p->left);
                }
                if(p->right != NULL) {
                    qu.push(p->right);
                }
            }
        }
        return ans;
    }
};
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运行结果

C语言BFS版实现代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
#define MAXSIZE 10003
int maxDepth(struct TreeNode* root) {
    if(root == NULL) {
        return NULL;
    }
    int front = 0, rear = 0;
    int len = 0;
    struct TreeNode** queue = (struct TreeNode**)malloc(sizeof(struct TreeNode *) * MAXSIZE);
    queue[rear++] = root;
    int ans = 0;
    while(front != rear) {
        len = rear - front;
        ans++;
        while(len > 0) {
            len--;
            struct TreeNode* tmp = queue[front++];
            if(tmp->left) {
                queue[rear++] = tmp->left;
            }
            if(tmp->right) {
                queue[rear++] = tmp->right;
            }
        }
    }
    return ans;
}
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运行结果

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