题目描述

我的思路很简单，就是简单的bfs算法，每次遍历完当前一层的节点，顺带将下一层的非nullptr节点加进来。循环截止的出口是队列为空。

算法1

(bfs) $O(n)$

C++ 代码

/**
 * 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 treeDepth(TreeNode* root) {
        queue<TreeNode*> q;
        int depth = 0;
        if(!root) return 0;
        q.push(root);
        while(q.size())
        {
            depth++;
            int n = q.size();
            for(int i = 0; i < n; i ++ )
            {
                TreeNode* t = q.front();
                q.pop();
                if(t->left) q.push(t->left);
                if(t->right) q.push(t->right);
            }
        }
        return depth;
    }
};