題目：

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.

思路：

題意是要從根幾點root到葉子節點的所有組合，是不是等於給定的整數sum 考慮用遞歸，把對當前的判斷，轉化為root.left和sum -root.val的比較以及toot.right和sum-root.val的比較 -

代碼：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null){
            return false;
        }
        int count = 0;
        if(root != null){
            count = root.val;
        }
            if(root.left != null && root.right != null){
                return hasPathSum(root.left,sum-count) || hasPathSum(root.right,sum-count);
            }
            if(root.right != null){
                return hasPathSum(root.right,sum-count);
            }
            if(root.left != null){
                return hasPathSum(root.left,sum-count);
            }
            if(count == sum){
                return true;
            }
            return false;
    }
}