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.
代码:
/**
* 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:
bool hasPathSum(TreeNode* root, int sum) {
if(!root)return false;
if(!root->left&&!root->right){
if(sum==root->val)return true;
else return false;
}
bool b1=false,b2=false;
int value=sum-root->val;
if(root->left){
b1=hasPathSum(root->left,value);
}
if(root->right){
b2=hasPathSum(root->right,value);
}
return b1||b2;
}
};
想法:
考虑极端情况