Leetcode/微软bloogberg -- 112.113.Path Sum I + II(dfs)

112.113. Path Sum I/II (Backtracking)
easy medium
https://leetcode.com/problems/path-sum/
https://leetcode.com/problems/path-sum-ii/
相关题 Path III，区别在于path可以树中任意路径上的节点，所以2次dfs，多的一次用来计算树中非root的pathSum http://rainykat.blogspot.com/2017/01/leetcode-437-path-sum-iii-2dfs.html

113.

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]
]

112只要返回有无路径,return true or false

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null)return false;
        if(root.left == null && root.right == null && sum - root.val == 0)
            return true;
        return hasPathSum(root.left, sum - root.val)||hasPathSum(root.right, sum - root.val);
    }
}

113需要存下path, back track path

public class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if(root == null)return res;
        helper(root, res, new ArrayList<Integer>(), sum);
        return res;
    }
    public void helper(TreeNode root, List<List<Integer>> res, List<Integer> path, int sum){
        if(root == null)return;
        path.add(root.val);
        if(root.left == null && root.right == null && sum - root.val == 0)
            res.add(new ArrayList<>(path));
        helper(root.left, res, path, sum - root.val);
        helper(root.right, res, path, sum - root.val);
        path.remove(path.size()-1);
    }
}