leetcode_112题——Path Sum（二叉树，深度优先搜索）

Path Sum

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Question Solution

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.

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这道题需要查找出是否有路径等于给定值，采用深度优先搜索的方法，在每次往下搜索时，记录下到当前结点的路径，然后若到了叶节点时

就判断下是否与之相等，再将判断的结果回溯就可以了：

#include<iostream>
#include<stack>
#include<set>

using namespace std;
// Definition for binary tree
struct TreeNode
{
	int val;
    TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

bool ifsum(TreeNode* root,int sum,int cursum)
{
	if(root==NULL)
		return false;
	if(root->left==NULL&&root->right==NULL)
		return cursum+root->val==sum;

	if(root->left!=NULL||root->right!=NULL)
		return ifsum(root->left,sum,cursum+root->val)||ifsum(root->right,sum,cursum+root->val);
}
bool hasPathSum(TreeNode *root, int sum) {
	return ifsum(root,sum,0);
}
int main()
{

}

　　还看到有一种基本与上面类似的方法，只是将当前的路径放在了外面作为迁居变量，而这里需要注意的是，在每次向下搜索完了之后，要记得在之前加了当前结点值的要减掉这个值

#include<iostream>
#include<stack>
#include<set>

using namespace std;
// Definition for binary tree
struct TreeNode
{
	int val;
    TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

int sum_temp;
int targetsum;
bool ifsum(TreeNode* root)
{
	if(root==NULL)
		return false;
	if(root->left==NULL&&root->right==NULL)
		return targetsum+root->val==sum_temp;

	targetsum+=root->val;
	bool res1=ifsum(root->left);
	bool res2=ifsum(root->right);

	targetsum-=root->val;//这里由于是全局变量，所以在前面加了之后，在这个结点算完后，要减掉

	return res1||res2;
}

bool hasPathSum(TreeNode *root, int sum) {
	sum_temp=sum;
	targetsum=0;
	return ifsum(root);
}
int main()
{

}