[Leetcode][JAVA] Path Sum I && II

Path Sum

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

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

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

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

分治法可以解决，传入下层的sum为原来的sum-root.val即可

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

 

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]
]

II就是要返回所有可能的path. 可以用分治法的思想去实现（把根节点加到左子树得到的list和右子树得到的list的第一位），不过较慢，因为要结果返回给上层。用单纯的dfs回溯也能很好地实现，而且较快。

分治法：

  public List<List<Integer>> pathSum(TreeNode root, int sum) {
         List<List<Integer>> re = new ArrayList<List<Integer>>();
         if(root==null)
             return re;
         if(root.left==null && root.right==null && root.val==sum) {
             List<Integer> temp = new ArrayList<Integer>();
             temp.add(root.val);
             re.add(temp);
             return re;
         }
         List<List<Integer>> left = pathSum(root.left, sum-root.val);
         List<List<Integer>> right = pathSum(root.right, sum-root.val);
         if(left.size()>0)
             for(int i=0;i<left.size();i++) {
                 left.get(i).add(0,root.val);
                 re.add(left.get(i));
             }
         if(right.size()>0)
             for(int i=0;i<right.size();i++) {
                 right.get(i).add(0,root.val);
                 re.add(right.get(i));
             }
         return re;
     }

dfs：

public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> re = new ArrayList<List<Integer>>();
        if(root==null)
            return re;
        List<Integer> path = new ArrayList<Integer>();
        collect(re, path, root, sum);
        return re;
    }

    public void collect(List<List<Integer>> re, List<Integer> path, TreeNode rt, int v) {
        if(rt.left==null && rt.right==null && rt.val==v) {
            List<Integer> temp = new ArrayList<Integer>(path);
            temp.add(rt.val);
            re.add(temp);
            return;
        }
        path.add(rt.val);
        if(rt.left!=null)
            collect(re,path,rt.left,v-rt.val);
        if(rt.right!=null)
            collect(re, path, rt.right, v-rt.val);
        path.remove(path.size()-1);
    }