Path Sum

【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.

【题意】

     判断二叉树是否存在路径和等于sum的路径，若存在输出true，否则输出false
【分析】

    递归调用二叉树，每次将上一层的val值传递给子结点并加上子节点的val，当传递到某个结点为叶子结点时，判断其val值是否等于sum

【实现】

     /**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {

return pathFind(root, sum);
}
public boolean pathFind(TreeNode root, int sum){
if (root == null){
return false;
}
if (root.left == null && root.right == null){
if( sum - root.val == 0){
return true;
}
}
return (pathFind(root.left, sum - root.val) || pathFind(root.right, sum - root.val));
}
}c++
/**
* 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 dfs(TreeNode *node, int sum, int curSum)
{
if (node == NULL)
return false;

if (node->left == NULL && node->right == NULL)
return curSum + node->val == sum;

return dfs(node->left, sum, curSum + node->val) || dfs(node->right, sum, curSum + node->val);
}

bool hasPathSum(TreeNode *root, int sum) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
return dfs(root, sum, 0);
}
};