Leetcode House Robber III

Source: Internet
Author: User

The original title link is here: https://leetcode.com/problems/house-robber-iii/


The thief has found himself a new place for his thievery again. There is a entrance to this area, called the "root." Besides the root, each house have one and only one parent house. After a tour, the Smart thief realized the ' all houses in this ' place forms a binary tree. It would automatically contact the police if and 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:

     3    /    2   3    \   \      3   1

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

Example 2:

     3    /    4   5  /\   \  1   3   1

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


Robroot = notrobleft + notrobright + root.val

Notrobroot = Math.max (Robleft, Notrobleft) + Math.max (Robright, Notrobright).

Dfs from the tree is updated from the bottom up.

Time Complexity:o (n). Space:o (LOGN). The LOGN layer stack is used.

AC Java:

1 /**2 * Definition for a binary tree node.3 * public class TreeNode {4 * int val;5 * TreeNode left;6 * TreeNode right;7 * TreeNode (int x) {val = x;}8  * }9  */Ten  Public classSolution { One      Public intRob (TreeNode root) { A         int[] res =Dfs (root); -         returnMath.max (Res[0], res[1]); -     } the     Private int[] DFS (TreeNode root) { -         int[] DP =New int[2]; -         if(Root = =NULL){ -             returnDP; +         } -         int[] left =DFS (root.left); +         int[] right =DFS (root.right); A         //dp[0] means stealing root, then can't steal around, so use left[1], right[1]. atDp[0] = left[1] + right[1] +Root.val; -         //dp[1] means not to steal root, then the right or left to steal can, take the maximum value -DP[1] = Math.max (left[0], left[1]) + Math.max (right[0], right[1]); -         returnDP; -     } -}

Similar to house robber, house robber Ii.

Leetcode House Robber III

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