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LeetCode 404 Sum of Left Leaves

2024-08-27 10:55:19   220人阅读

Problem:

Find the sum of all left leaves in a given binary tree.

Summary:

求左子叶之和。

Analysis：

1. 求左子叶之和，即需要遍历整棵二叉树，常规的方法则为递归。首先我想到的是调用子函数，在子函数中递归，每递归至一个新的节点，首先传入bool类型标记是根节点的左子树还是右子树。若为左子树，且为叶子节点，则将当前节点的value计入和中。

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int sumOfLeftLeaves(TreeNode* root) {
13         if (root == NULL) {
14             return 0;
15         }
16         
17         int sum = 0;
18         calLeftLeaves(root->left, true, sum);
19         calLeftLeaves(root->right, false, sum);
20         
21         return sum;
22     }
23 
24 private:
25     void calLeftLeaves(TreeNode* root, bool left, int& sum) {
26         if (root == NULL) {
27             return;
28         }
29         
30         if (left && (root->left == NULL) && (root->right == NULL)) {
31             sum += root->val;
32         }
33         
34         calLeftLeaves(root->left, true, sum);
35         calLeftLeaves(root->right, false, sum);
36     }
37 };

 2. 进一步简化代码，试图不掉用子函数，在主函数中进行递归操作。

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int sumOfLeftLeaves(TreeNode* root) {
13         if (root == NULL) {
14             return 0;
15         }
16         
17         if (root->left && root->left->left == NULL && root->left->right == NULL) {
18             return root->left->val + sumOfLeftLeaves(root->right);
19         }
20         
21         return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
22     }
23 };

 

LeetCode 404 Sum of Left Leaves


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