Path Sum II

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]
]

Analysis

DFS problem

Code

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>();
        dfs(root, sum, list, result);
        return result;
    }

    private void dfs(TreeNode root, int sum, List<Integer> list, List<List<Integer>> result){
        if (root == null) {return;}
        list.add(root.val);
        if (root.left == null && root.right == null && root.val == sum){
            result.add(new ArrayList<Integer>(list)); 
            //IMPORTANT!! Easy to forget!
            list.remove(list.size()-1);
            return;
        }
        dfs(root.left, sum-root.val, list, result);
        dfs(root.right, sum-root.val, list, result);
        list.remove(list.size()-1);
    }
}

Note

1. 在code中把list加入到result后，注意要remove最后一个element。注意在return前要把之前加入list的删除。
2. result加入list记住要copy一个新的list加入，不然之后的操作会改变list的值。
3. list可以加入的条件是到达leaf node并且sum为target

Reference

Ref1