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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 tree 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->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：
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原文地址：http://www.cnblogs.com/wuchanming/p/4095944.html
