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leetcode 404 Sum of Left Leaves
Find the sum of all left leaves in a given binary tree.
Example:
3 / 9 20 / 15 7There 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 else19 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 };
leetcode 404 Sum of Left Leaves
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