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 };