404. Sum of Left Leaves
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Problem:
Solution:
看到树,必然想到要用递归算法,本题的递归方式为:
1、若root为空,则返回0;
2、若root的左子树root->left为叶子结点,此时的root->left正是我们需要计算其值的目标结点,返回root->left->val + 对root的右子树的递归访问
3、若为其他情况,则返回对root的左右子树分别的递归访问之和
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: int sumOfLeftLeaves(TreeNode* root) { if (root == NULL) { return 0; } if (root->left != NULL && root->left->left == NULL && root->left->right == NULL) { return root->left->val + sumOfLeftLeaves(root->right); } else return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right); }};
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- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves*
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
- 404. Sum of Left Leaves
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