【Leetcode】Path sum 2

【題目】

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]
]

【思路】

Save intermediate result into stack and save the stack into result array once its sum == required sum.

注意一定是要從根到葉子。所以最後要確定左右節點都是空。

list可以直接添加stack?

【代碼】

public class Solution {
    private List<List<Integer>> resultList = new ArrayList<List<Integer>>();

    public void pathSumInner(TreeNode root, int sum, Stack<Integer>path) {
        path.push(root.val);
        if(root.left == null && root.right == null)
            if(sum == root.val) resultList.add(new ArrayList<Integer>(path));
        if(root.left!=null) pathSumInner(root.left, sum-root.val, path);
        if(root.right!=null)pathSumInner(root.right, sum-root.val, path);
        path.pop();
    }

    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        if(root==null) return resultList;
        Stack<Integer> path = new Stack<Integer>();
        pathSumInner(root, sum, path);
        return resultList;
    }
}