[LintCode] 二叉树的路径之和 Binary Tree Path Sum

给定一个二叉树，找出所有路径中各节点相加总和等于给定 目标值 的路径。

一个有效的路径，指的是从根节点到叶节点的路径。

样例

给定一个二叉树，和 目标值 = 5:

1

/ \

2 4

/ \

2 3

返回：

[

[1, 2, 2],

[1, 4]

]

Given a binary tree, find all paths that sum of the nodes in the path equals to a given number target.

A valid path is from root node to any of the leaf nodes.

/**
* Definition of TreeNode:
* public class TreeNode {
*     public int val;
*     public TreeNode left, right;
*     public TreeNode(int val) {
*         this.val = val;
*         this.left = this.right = null;
*     }
* }
*/
public class Solution {
/**
* @param root the root of binary tree
* @param target an integer
* @return all valid paths
*/
public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if(null == root) return result;
Stack<Integer> stack = new Stack<Integer>();
binaryTreePathSum(result, stack, root, 0, target);
return result;
}

public void binaryTreePathSum(List<List<Integer>> result, Stack<Integer> stack, TreeNode root, int sum, int target) {
sum += root.val;
stack.push(root.val);
if(sum == target && root.left == null && root.right == null) {
List<Integer> list = new ArrayList<Integer>(stack);
result.add(list);
stack.pop();
return;
}else {
if(root.left != null)
binaryTreePathSum(result, stack, root.left, sum, target);
if(root.right != null)
binaryTreePathSum(result, stack, root.right, sum, target);
stack.pop();
}
}
}