原体如下：
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.
For example:Given the below binary tree andsum = 22,5/ \4 8/ / \11 13 4/ \\7 21
return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.
我的代码：
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();}}
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，最大的成功在于最大的付出。
