112. Path Sum LeetCode
2016-02-11 23:02
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题意:给一棵二叉树和一个sum,问是否存在一条从根到叶子的路径,把经过的节点的数字加和等于sum。
题解:递归求解,左儿子不空,那么递归判断左子树是否存在路径使得加和等于sum - root->val。右儿子不空,那么递归判断右子树是否存在路径使得加和等于sum - root->val.
题解:递归求解,左儿子不空,那么递归判断左子树是否存在路径使得加和等于sum - root->val。右儿子不空,那么递归判断右子树是否存在路径使得加和等于sum - root->val.
class Solution { public: bool hasPathSum(TreeNode* root, int sum) { if(root == NULL) return false; if(root->left != NULL && root->right != NULL) return hasPathSum(root->left,sum - root->val) || hasPathSum(root->right,sum - root->val); else if(root->left == NULL && root->right != NULL) return hasPathSum(root->right,sum - root->val); else if(root->left != NULL && root->right == NULL) return hasPathSum(root->left,sum - root->val); else return !(sum - root->val); } };
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