[LeetCode]112.Path Sum
栏目：php教程时间：2015-01-04 08:56:25
【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.
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.
【分析】
题目要求是从根节点到叶子节点的1条路径。刚开始没看清楚，还以为随便路径。
题目只要求返回true或false，因此没有必要记录路径。
【代码】
/*********************************
* 日期：2015-01-01
* 作者：SJF0115
* 题目: 112.Path Sum
* 来源：https://oj.leetcode.com/problems/path-sum/
* 结果：AC
* 来源：LeetCode
* 博客：
* 时间复杂度：O(n)
* 空间复杂度：O(logn)
**********************************/
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
// 2叉树节点
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 == NULL){
return false;
}//if
// 找到1条从根节点到叶子节点的路径
if(root->left == NULL && root->right == NULL){
return sum == root->val;
}//if
// 左子树
bool left = hasPathSum(root->left,sum - root->val);
// 右子树
bool right = hasPathSum(root->right,sum - root->val);
return left || right;
}
};
// 创建2叉树
TreeNode* CreateTreeByLevel(vector<int> num){
int len = num.size();
if(len == 0){
return NULL;
}//if
queue<TreeNode*> queue;
int index = 0;
// 创建根节点
TreeNode *root = new TreeNode(num[index++]);
// 入队列
queue.push(root);
TreeNode *p = NULL;
while(!queue.empty() && index < len){
// 出队列
p = queue.front();
queue.pop();
// 左节点
if(index < len && num[index] != ⑴){
// 如果不空创建1个节点
TreeNode *leftNode = new TreeNode(num[index]);
p->left = leftNode;
queue.push(leftNode);
}
index++;
// 右节点
if(index < len && num[index] != ⑴){
// 如果不空创建1个节点
TreeNode *rightNode = new TreeNode(num[index]);
p->right = rightNode;
queue.push(rightNode);
}
index++;
}//while
return root;
}
int main() {
Solution solution;
// ⑴代表NULL
vector<int> num = {5,4,8,11,⑴,13,4,7,2,⑴,⑴,⑴,1};
TreeNode* root = CreateTreeByLevel(num);
cout<<solution.hasPathSum(root,22)<<endl;
}
------分隔线----------------------------
------分隔线----------------------------