[数据结构与算法]【LeetCode-222 | 完全二叉树的节点个数】

#include <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>

using namespace std;


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) {}
};


/*
    递归法：使用前序遍历计算完全二叉树的节点个数，时间复杂度O(n)，空间复杂度O(logn)
*/
class Solution {
public:
    // 1.确定函数入参及返回值
    int getNodesNum(TreeNode* node) {
        // 2.确定递归的终止条件
        if(node == nullptr) return 0;

        // 3.确定单层递归逻辑
        int leftNum = getNodesNum(node->left);
        int rightNum = getNodesNum(node->right);

        int result = 1 + leftNum + rightNum;
        return result; 
    }

    int countNodes(TreeNode* root) {
        return getNodesNum(root);
    }
};

/*
    迭代法：使用层序遍历计算完全二叉树的节点个数，时间/空间复杂度O(n)
*/
class Solution {
public:
    int getNodesNum(TreeNode* root) {
        queue<TreeNode*> que;
        if(root != nullptr) que.push(root);

        int result = 0;
        while(!que.empty()) {
            int size = que.size();
            for(int i = 0; i < size; i++) {
                TreeNode* cur = que.front();
                que.pop();
                result++;

                if(cur->left) que.push(cur->left);
                if(cur->right) que.push(cur->right);
            }
        }
        return result;
    }
};