LeetCode – House Robber III (Java)

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The houses form a binary tree. If the root is robbed, its left and right can not be robbed.

Analysis

Traverse down the tree recursively. We can use an array to keep 2 values: the maximum money when a root is selected and the maximum value when a root if NOT selected.

Java Solution

public int rob(TreeNode root) {
    if(root == null)
        return 0;
 
    int[] result = helper(root);
    return Math.max(result[0], result[1]);
}
 
public int[] helper(TreeNode root){
    if(root == null){
        int[] result = {0, 0};
        return result;
    }
 
    int[] result = new int[2];
    int[] left = helper(root.left);
    int[] right = helper (root.right);
 
    // result[0] is when root is selected, result[1] is when not. 
    result[0] = root.val + left[1] + right[1];
    result[1] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
 
    return result;
}

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