337 House Robber III

337. House Robber III

1. Question

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:
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3
2
3
/ \
4
2 3
5
\ \
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3 1
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Maximum amount of money the thief can rob =
3 + 3 + 1 = 7.
Example 2:
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3
2
/ \
3
4 5
4
/ \ \
5
1 3 1
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Maximum amount of money the thief can rob =
4 + 5 = 9.

2. Implementation

(1) DFS
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public class Solution {
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public int rob(TreeNode root) {
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Map<TreeNode, Integer> map = new HashMap<>();
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return robHouse(root, map);
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}
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public int robHouse(TreeNode node, Map<TreeNode, Integer> map) {
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if (node == null) {
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return 0;
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}
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if (map.containsKey(node)) {
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return map.get(node);
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}
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int val = 0;
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if (node.left != null) {
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val += robHouse(node.left.left, map) + robHouse(node.left.right, map);
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}
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if (node.right != null) {
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val += robHouse(node.right.left, map) + robHouse(node.right.right, map);
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}
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int res = Math.max(val + node.val, robHouse(node.left, map) + robHouse(node.right, map));
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map.put(node, res);
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return res;
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}
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}
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3. Time & Space Complexity

DFS: 时间复杂度O(n), 空间复杂度O(n),因为用了map