LeetCode OJ

二叉树最经典的解法就是采用递归方法，这道题也不例外。下面是AC代码：

 /**
     * 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.
     * @param root
     * @param sum
     * @return
     */
    public boolean hasPathSum(TreeNode root, int sum){
        if(root == null )
            return false;
        return recurPathSum(root,sum,0);
    }
    private boolean recurPathSum(TreeNode root, int sum , int curr){
        if(root == null)
            return sum == curr;
        if(root.left == null && root.right == null)
        {
            return curr+root.val==sum;
        }
        return (root.left == null ? false :recurPathSum(root.left,sum, curr+root.val)) || 
                (root.right==null ? false : recurPathSum(root.right, sum, curr+root.val));
    }
    /**
     * Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
     * @param root
     * @param sum
     * @return
     */
    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum){
        ArrayList<ArrayList<Integer>> r = new ArrayList<ArrayList<Integer>>();
        if(root == null)
            return r;
        recurPathSum(root, sum, 0, new ArrayList<Integer>(), r);
        return r;
    }
    private void recurPathSum(TreeNode root,int sum, int curr, 
            ArrayList<Integer> path, ArrayList<ArrayList<Integer>>paths){
        if(root == null){
            if(curr == sum)
                paths.add(path);
            return;
        }
        if(root.left == null && root.right == null){
            if(curr + root.val == sum)
            {
                path.add(root.val);
                paths.add(path);
            }
            return;
        }
        if(root.left != null)
        {
            ArrayList<Integer> p = new ArrayList<Integer>();
            p.addAll(path);
            p.add(root.val);
            recurPathSum(root.left,sum,curr+root.val,p,paths);
        }
        if(root.right != null)
        {
            ArrayList<Integer> p = new ArrayList<Integer>();
            p.addAll(path);
            p.add(root.val);
            recurPathSum(root.right,sum,curr+root.val,p,paths);
        }
    }