1. 程式人生 > 實用技巧 >2020牛客多校第七場C-A National Pandemic

2020牛客多校第七場C-A National Pandemic

題目描述

輸入一棵二叉樹的根節點,求該樹的深度。從根節點到葉節點依次經過的節點(含根、葉節點)形成樹的一條路徑,最長路徑的長度為樹的深度。

例如:
給定二叉樹 [3,9,20,null,null,15,7]

    3
   / \
  9  20
    /  \
   15   7

返回它的最大深度 3 。

提示:

節點總數 <= 10000

來源:力扣(LeetCode)
連結:https://leetcode-cn.com/problems/er-cha-shu-de-shen-du-lcof

程式碼實現

/**
 * 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) {
        if(!root)   return 0;
        queue<TreeNode*> node_queue;
        TreeNode* p = root;
        node_queue.push(p);
        int layer_num = 0;
        while(!node_queue.empty()) {
            queue<TreeNode*> tmp;
            while(!node_queue.empty()) {
                TreeNode* cur_node = node_queue.front();
                if(cur_node->left)
                    tmp.push(cur_node->left);
                if(cur_node->right)
                    tmp.push(cur_node->right);
                node_queue.pop();
            }
            node_queue = tmp;
            layer_num++;
        } 
        return layer_num;
    }
};

思路解析

  • 層序遍歷二叉樹,使用tmp佇列儲存當前層的所有節點,layer_num++