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,

      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：標準的考察二叉樹遍歷題目，深度優先遍歷，沒什麼好說的。

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