leetcode 404 Sum of Left Leaves

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mb5fe32930661bd 2017-02-18 15:51:00

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

Example:

    3
   / \
  9  20
    /  \
   15   7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.

分析：关键是怎么判断它是左叶子；

 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         TreeNode* temp = root->left;
16         if (temp && (temp->left == NULL) && (temp->right == NULL))
17             return temp->val + sumOfLeftLeaves(root->right);
18         else
19             return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
20     }
21 };

也可用bfs

网上大神的dfs：深度优先遍历，将所有结点从根结点开始遍历一遍，设立isLeft的值，当当前结点是叶子节点并且也是左边，那就result加上它的值

 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 result = 0;
13     int sumOfLeftLeaves(TreeNode* root) {
14         if(root == NULL)
15             return 0;
16         dfs(root, false);
17         return result;
18     }
19     void dfs(TreeNode* root, bool isLeft) {
20         if(root->left == NULL && root->right == NULL) {
21             if(isLeft == true)
22                 result += root->val;
23             return ;
24         }
25         if(root->left != NULL)
26             dfs(root->left, true);
27         if(root->right != NULL)
28             dfs(root->right, false);
29     }
30 };