House Robber III

问题：

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all houses in this place forms a binary tree". It will automatically contact the police if two directly-linked houses were broken into on the same night.

Determine the maximum amount of money the thief can rob tonight without alerting the police.

这个问题可以分解为子问题，因为对于二叉树其左右子树也是二叉树，我们定义一个函数rob(root)让它返回根节点开始小偷能偷到的最大金钱数额，我们可以由rob(root.left)和rob(root.right)来计算rob(root).所以原问题可以递归求解：

<span style="font-size:14px;">public int rob(TreeNode root) {    if (root == null) {        return 0;    }    int val = 0;    if (root.left != null) {        val += rob(root.left.left) + rob(root.left.right);    }    if (root.right != null) {        val += rob(root.right.left) + rob(root.right.right);    }    return Math.max(val + root.val, rob(root.left) + rob(root.right));}</span>

然而这样做很慢。因为深度越深的节点多次做重复计算，因此考虑用一个hash表存储每个节点能偷到的最大金额，从而问题变为一个DP问题。

<span style="font-size:14px;">public int rob(TreeNode root) {    Map<TreeNode, Integer> map = new HashMap<>();    return robSub(root, map);}private int robSub(TreeNode root, Map<TreeNode, Integer> map) {    if (root == null) return 0;    if (map.containsKey(root)) return map.get(root);    int val = 0;    if (root.left != null) {        val += robSub(root.left.left, map) + robSub(root.left.right, map);    }    if (root.right != null) {        val += robSub(root.right.left, map) + robSub(root.right.right, map);    }    val = Math.max(val + root.val, robSub(root.left, map) + robSub(root.right, map));    map.put(root, val);    return val;}</span>

然而这是一个O(n)空间复杂度，然而这个复杂度还可以降低，对于每个节点有偷或者不偷两种情况，我们可以给每个节点两个数表示偷和不偷分别获得的最大金额，而父节点的两个数又可以根据子节点的两个数得到，所以我们把空间复杂度可以降低为O（1）：

<span style="font-size:14px;">public int rob(TreeNode root) {    int[] res = robSub(root);    return Math.max(res[0], res[1]);}private int[] robSub(TreeNode root) {    if (root == null) {        return new int[2];    }    int[] left = robSub(root.left);    int[] right = robSub(root.right);    int[] res = new int[2];    res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);    res[1] = root.val + left[0] + right[0];    return res;}</span>

下面是C++版本：

<span style="font-size:14px;">rob(TreeNode* root) {    vector<int> res = robSub(root);    return max(res[0], res[1]);}vector<int> robSub(TreeNode* root) {    if (root == NULL) {        return vector<int>(2,0);    }    vector<int> left = robSub(root->left);    vector<int> right = robSub(root->right);    vector<int> res(2,0);    res[0] = max(left[0], left[1]) + max(right[0], right[1]);    res[1] = root->val + left[0] + right[0];    return res;}</span>