【Leetcode笔记】404.左叶子之和

这么久过去了，递归仍然是压4我的一座大山啊啊啊啊啊~~~~~

---

#include <iostream>
#include <vector>
#include <string>
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) {}
 };
class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) 
    {
        if(root == NULL) return 0;
        if(root->left == NULL && root->right == NULL) return 0;

        int left = sumOfLeftLeaves(root->left);
        if(root->left != NULL && root->left->left == NULL && root->left->right == NULL)
        {
            left = root->left->val;
        }
        int right = sumOfLeftLeaves(root->right);

        int sum = left + right;
        return sum;
    }
};

int main() {
    TreeNode* root = new TreeNode(3);
    root->left = new TreeNode(9);
    root->right = new TreeNode(20);
    root->right->left = new TreeNode(15);
    root->right->right = new TreeNode(7);
    Solution solution;
    int result = solution.sumOfLeftLeaves(root);
    cout << result << endl;
    return 0;
}

---

评论区看到一条前序遍历的解法，更容易理解：

class Solution {
    int res=0;//全局变量存放累加的左叶子节点之和
    public int sumOfLeftLeaves(TreeNode root) {
        if(root==null)return 0;//终止条件
       //满足下列条件，该节点的左子节点即是左叶子节点
        if(root.left!=null&&root.left.left==null&&root.left.right==null){
            res+=root.left.val;//判断左叶子节点只能从当前节点判断，如果是左叶子节点，加入结果集
        }
        sumOfLeftLeaves(root.left);//遍历左树
        sumOfLeftLeaves(root.right);//遍历右树
        return res;//遍历完所有节点，返回左叶子之和
    }
};