LeetCode[337] House Robber III

Dr_Noooo 發(fā)布于2019-08-14 15:00 / 919人閱讀

摘要：復(fù)雜度思路對(duì)于每一個(gè)位置來(lái)說(shuō)，考慮兩種情況分別對(duì)和再進(jìn)行計(jì)算。用對(duì)已經(jīng)計(jì)算過(guò)的進(jìn)行保留，避免重復(fù)計(jì)算。

LeetCode[337] House Robber III

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.

Example 1:

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

Maximum amount of money the thief can rob = 4 + 5 = 9.

recursion + Memorization

復(fù)雜度
O(N), O(lgN)

思路
對(duì)于每一個(gè)位置來(lái)說(shuō)，考慮兩種情況, Max(child, subchild + root.val).
分別對(duì)child和subchild再進(jìn)行recursion計(jì)算。

用map對(duì)已經(jīng)計(jì)算過(guò)的node進(jìn)行保留，避免重復(fù)計(jì)算。

代碼

public int rob(TreeNode root) {
    // Base case;
    if(root == null) return 0;
    if(root.left == null && root.right == null) return root.val;
    if(map.containsKey(root)) return map.get(root);
    int child = 0, subchild = 0;
    if(root.left != null) {
        child += rob(root.left);
        subchild += rob(root.left.left) + rob(root.left.right);
    }
    if(root.right != null) {
        child += rob(root.right);
        subchild += rob(root.right.left) + rob(root.right.right);
    }
    int val = Math.max(child, subchild + root.val);
    map.put(root, val);
    return val;
}