原体例如以下：

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.

              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.

public class Solution {
    public boolean mark = false;
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
			return false;
		}
		Stack<Integer> path = new Stack<Integer>();
		int currentSum = 0;
		findPath(root, sum, path, currentSum);
		return mark;
    }
    public void findPath(TreeNode root, int sum, Stack<Integer> path, int currentSum)
	{
		currentSum += root.val;
		path.push(root.val);
		boolean isLeaf = root.left==null&&root.right==null;
		if(currentSum==sum&&isLeaf)
		{
			mark = true;
		}
		if(root.left != null)
		{
			findPath(root.left,sum,path,currentSum);
		}
		if(root.right != null)
		{
			findPath(root.right,sum,path,currentSum);
		}
		
		path.pop();
	}
}