[LeetCode]112. Path Sum&113. Path Sum II

112 . Path Sum
Easy

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.

1ms:

 public boolean hasPathSum(TreeNode root, int sum) {         if(root==null)  return false;         return subPath(sum,0,root);     }     private boolean subPath(int sum,int v,TreeNode node){         v += node.val;         if(node.left==null&&node.right==null){             if(v==sum)                 return true;         }         if(node.left!=null){             if(subPath(sum,v,node.left))                 return true;         }         if(node.right!=null){             if(subPath(sum,v,node.right))                 return true;         }         return false;     }

0ms:

public boolean hasPathSum2(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);     }

113 . Path Sum II
Medium

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]
]

3ms:

public List<List<Integer>> pathSum(TreeNode root, int sum) {        List<List<Integer>> result = new ArrayList<List<Integer>>();        if(root==null) return result;        List<Integer> link = new LinkedList<Integer>();        subSum(sum,result,link,root);        return result;    }    private void subSum(int v,List<List<Integer>> result,List<Integer> list,TreeNode node){        list.add(node.val);        if(node.left==null&&node.right==null){            if(node.val==v)                result.add(new ArrayList<Integer>(list));        }        if(node.left!=null)            subSum(v-node.val,result,list,node.left);        if(node.right!=null)            subSum(v-node.val,result,list,node.right);        list.remove(list.size()-1);    }

参考：
257. Binary Tree Paths