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.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

Solution:

#include <iostream>

using namespace std;


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 ( root ->left == NULL && root ->right == NULL )
            return root ->val == 0;

        if ( hasPathSum( root->left , sum - root->val ))
            return true;

        if ( hasPathSum( root->right, sum - root->val ))
            return true;

        return false;

    }
};
 

總結： 要注意遞迴終止條件的設計。