Path Sum II

题目描述:

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

这个题比较简单，到了根节点的时候判断和是否等于sum，等于就将集合加入result即可。

代码如下：

public class Solution {
     public List<List<Integer>> pathSum(TreeNode root, int sum) {
		 List<List<Integer>> result=new ArrayList<List<Integer>>();
		 if(root==null)
			 return result;
		 List<Integer> list=new ArrayList<Integer>();
		 getPath(root, 0, sum, list, result);
		 return result;
	 }
	 
	 public void getPath(TreeNode root,int cursum,int sum,List<Integer> list,List<List<Integer>> result){
		 if(root.left==null&&root.right==null){
			 if(root.val+cursum==sum){
				 List<Integer> newList=new ArrayList<Integer>(list);
				 newList.add(root.val);
				 result.add(newList);
			 }
		 }
		 if(root.left!=null){
			 List<Integer> newList=new ArrayList<Integer>(list);
			 newList.add(root.val);
			 getPath(root.left, cursum+root.val, sum, newList, result);
		 }
		 if(root.right!=null){
			 List<Integer> newList=new ArrayList<Integer>(list);
			 newList.add(root.val);
			 getPath(root.right, cursum+root.val, sum, newList, result);
		 }
	 }
}