404. Sum of Left Leaves

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.

求左叶子结点的值之和

java（10ms）：
 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public int sumOfLeftLeaves(TreeNode root) {
12         if (root == null)
13         return 0  ;
14         int ans = 0 ;
15         Queue<TreeNode> que = new LinkedList<TreeNode>() ;
16         que.offer(root) ;
17         while(!que.isEmpty()){
18             TreeNode cur = que.poll() ;
19             if (cur.left != null && cur.left.left==null && cur.left.right == null)
20                ans += cur.left.val ;
21             if (cur.left != null)
22                 que.offer(cur.left) ;
23             if (cur.right != null)
24                 que.offer(cur.right) ;
25         }
26         return ans ;
27     }
28 }

 

C++(6ms):

 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         queue<TreeNode*> q ;
16         q.push(root) ;
17         int sum = 0 ;
18         while(!q.empty()){
19             TreeNode* t = q.front();
20             q.pop();
21             if(t->left != NULL && t->left->left == NULL && t->left->right == NULL)
22                 sum += t->left->val ;
23             if(t->left != NULL)
24                q.push(t->left) ;
25             if(t->right != NULL)
26                q.push(t->right) ;
27 
28         }
29         return sum ;
30     }
31 };

 

C++:

 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         queue<TreeNode*> q ;
16         q.push(root) ;
17         int sum = 0 ;
18         while(!q.empty()){
19             int len  = q.size() ;
20             for(int i = 0 ; i < len ; i++){
21                 TreeNode* t = q.front();
22                 q.pop();
23                 if(t->left != NULL && t->left->left == NULL && t->left->right == NULL)
24                     sum += t->left->val ;
25                 if(t->left != NULL)
26                    q.push(t->left) ;
27                 if(t->right != NULL)
28                    q.push(t->right) ;
29             }
30         }
31         return sum ;
32     }
33 };

 

posted @ 2017-03-07 14:39  __Meng  阅读(93)  评论(0)  编辑  收藏  举报