【LeetCode】Path Sum II 二叉树递归
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]
]
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
if(root == null)
return res;
ArrayList<Integer> tmp = new ArrayList<Integer>();
recursion(root, res, tmp, sum);
tmp = null;
return res;
}
public void recursion(TreeNode root, ArrayList<ArrayList<Integer>> res, ArrayList<Integer> tmp, int sum)
{
if(root.left == null && root.right == null)
{
tmp.add(root.val);
int sum1 = 0;
for(int i = 0; i < tmp.size(); i++)
{
sum1 += tmp.get(i);
}
if(sum1 == sum)
res.add(new ArrayList<Integer>(tmp));
return ;
}
tmp.add(root.val);
if(root.left != null)
{
recursion(root.left, res, tmp, sum);
tmp.remove(tmp.size() - 1);
}
if(root.right != null)
{
recursion(root.right, res, tmp, sum);
tmp.remove(tmp.size() - 1);
}
}
}
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