# Path-Sum

Nov 29, 2017

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example: Given the below binary tree and sum = 22,

          5
/ \
4   8
/   / \
11  13  4
/  \      \
7    2      1


return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

• 它要求一定要到left.
• 然后空节点不能当成0.

bool hasPathSum(TreeNode* root, int sum) {
if (!root ) return false;
if (!root->left && !root->right) return sum == root->val;
if (!root->left) return hasPathSum(root->right,sum-root->val);
if (!root->right) return hasPathSum(root->left,sum - root->val);
return hasPathSum(root->left,sum - root->val) || hasPathSum(root->right,sum - root->val);
}


bool hasPathSum(TreeNode *root, int sum) {
if (root == NULL) return false;
if (root->val == sum && root->left ==  NULL && root->right == NULL) return true;
return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);
}

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