题目
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.
Note: A leaf is a node with no children.
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.
解析
该题有点像按图索骥,从root开始按照前序遍历的方式进行,可以使用栈来做,但使用递归更加方便一些。
使用递归即是让“索骥”交给下一层去做,其sum也要相应减去根结点的值,这样到最底层的时候,若val == sum,那就是发现了一条路径,如果没有就返回false。
代码(C语言)
bool hasPathSum(struct TreeNode* root, int sum) {
if (root == NULL)
return false;
if (root->left == NULL && root->right == NULL && root->val == sum)
return true;
return hasPathSum(root->left, sum - root->val) ||
hasPathSum(root->right, sum - root->val);
}
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