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Path Sum ****
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.
C++代码：
#include<iostream>#include<new>#include<vector>using namespace std;//Definition for binary treestruct 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->left==NULL&&root->right==NULL&&(sum-root->val)==0) return true; if(root->left) return hasPathSum(root->left,sum-root->val); if(root->right) return hasPathSum(root->right,sum-root->val); return false; } void createTree(TreeNode *&root) { int i; cin>>i; if(i!=0) { root=new TreeNode(i); if(root==NULL) return; createTree(root->left); createTree(root->right); } }};int main(){ Solution s; TreeNode *root; s.createTree(root); if(s.hasPathSum(root,6)) cout<<"exist"<<endl;}
运行结果：
Path Sum ****
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