【算法与数据结构】404、LeetCode左叶子之和

所有的LeetCode题解索引，可以看这篇文章——【算法和数据结构】LeetCode题解。

一、题目

二、解法

  思路分析：思路比较简单，遍历所有节点然后判断该节点是否为左叶子节点，如果是，将其值累加。遍历算法采用【算法与数据结构】144、94、145LeetCode二叉树的前中后遍历（递归法、迭代法）文章中递归前序遍历算法，设置了一个左节点标志符，遍历左节点时，标识符为1，左右节点为空则为叶子节点。
  程序如下：

class Solution {
public:void traversal_preOrder(TreeNode* cur, int& sum, int left_flag) {  // 前序遍历if (cur == NULL) return;if (!cur->left && !cur->right && left_flag) sum += cur->val;  // 叶子节点且为左节点traversal_preOrder(cur->left, sum, 1);      // 左traversal_preOrder(cur->right, sum, 0);     // 右   }// 遍历所有节点，判断每个节点的左节点是否为叶子结点，然后相加int sumOfLeftLeaves(TreeNode* root) {       int result = 0;traversal_preOrder(root, result, 0);return result;}
};

复杂度分析：

• 时间复杂度： $O(n)$。
• 空间复杂度：$O(n)$。

三、完整代码

# include<iostream>
# include<string>
# include<vector>
# include<queue>
using namespace std;// 树节点定义
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:void traversal_preOrder(TreeNode* cur, int& sum, int left_flag) {  // 前序遍历if (cur == NULL) return;if (!cur->left && !cur->right && left_flag) sum += cur->val;  // 叶子节点且为左节点traversal_preOrder(cur->left, sum, 1);      // 左traversal_preOrder(cur->right, sum, 0);     // 右   }// 遍历所有节点，判断每个节点的左节点是否为叶子结点，然后相加int sumOfLeftLeaves(TreeNode* root) {       int result = 0;traversal_preOrder(root, result, 0);return result;}
};// 前序遍历迭代法创建二叉树，每次迭代将容器首元素弹出（弹出代码还可以再优化）
void Tree_Generator(vector<string>& t, TreeNode*& node) {if (!t.size() || t[0] == "NULL") return;    // 退出条件else {node = new TreeNode(stoi(t[0].c_str()));    // 中if (t.size()) {t.assign(t.begin() + 1, t.end());Tree_Generator(t, node->left);              // 左}if (t.size()) {t.assign(t.begin() + 1, t.end());Tree_Generator(t, node->right);             // 右}}
}// 层序遍历
vector<vector<int>> levelOrder(TreeNode* root) {queue<TreeNode*> que;if (root != NULL) que.push(root);vector<vector<int>> result;while (!que.empty()) {int size = que.size();  // size必须固定, que.size()是不断变化的vector<int> vec;for (int i = 0; i < size; ++i) {TreeNode* node = que.front();que.pop();vec.push_back(node->val);if (node->left) que.push(node->left);if (node->right) que.push(node->right);}result.push_back(vec);}return result;
}template<typename T>
void my_print(T& v, const string msg)
{cout << msg << endl;for (class T::iterator it = v.begin(); it != v.end(); it++) {cout << *it << ' ';}cout << endl;
}template<class T1, class T2>
void my_print2(T1& v, const string str) {cout << str << endl;for (class T1::iterator vit = v.begin(); vit < v.end(); ++vit) {for (class T2::iterator it = (*vit).begin(); it < (*vit).end(); ++it) {cout << *it << ' ';}cout << endl;}
}int main()
{vector<string> t = { "3", "9", "NULL", "NULL", "20", "15", "NULL", "NULL", "7", "NULL", "NULL" };   // 前序遍历my_print(t, "目标树");TreeNode* root = new TreeNode();Tree_Generator(t, root);vector<vector<int>> tree = levelOrder(root);my_print2<vector<vector<int>>, vector<int>>(tree, "目标树:");Solution s;int result = s.sumOfLeftLeaves(root);cout << "sumOfLeftLeaves:  " << result << endl;system("pause");return 0;
}

end