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:
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.

 /*************************************************************************
     > File Name: LeetCode337.c
     > Author: Juntaran
     > Mail: JuntaranMail@gmail.com
     > Created Time: Wed 11 May 2016 19:08:25 PM CST
  ************************************************************************/

 /*************************************************************************

     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:
          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.

  ************************************************************************/

 #include <stdio.h>

 /*
     继198与213
 */

 /*
     Discuss区大神答案
     https://leetcode.com/discuss/91777/intuitive-still-efficient-solution-accepted-well-explained
     另外有C++详解
     https://leetcode.com/discuss/91899/step-by-step-tackling-of-the-problem
 */

 /**
  * Definition for a binary tree node.
  * struct TreeNode {
  *     int val;
  *     struct TreeNode *left;
  *     struct TreeNode *right;
  * };
  */

 #define MAX(a, b) ((a) > (b) ? (a) : (b))

 // struct TreeNode
 // {
     // int val;
     // struct TreeNode *left;
     // struct TreeNode *right;
 // };

 void traverse( struct TreeNode* root, int* maxWithRoot, int* maxWithoutRoot )
 {
     int leftMaxWithRoot  = , leftMaxWithoutRoot  = ;
     int rightMaxWithRoot = , rightMaxWithoutRoot = ;

     if( root )
     {
         traverse( root->left,  &leftMaxWithRoot,  &leftMaxWithoutRoot  );
         traverse( root->right, &rightMaxWithRoot, &rightMaxWithoutRoot );

         *maxWithRoot    = leftMaxWithoutRoot + rightMaxWithoutRoot + root->val;
         *maxWithoutRoot = MAX( leftMaxWithRoot, leftMaxWithoutRoot ) + MAX( rightMaxWithRoot, rightMaxWithoutRoot );
     }
 }

 int rob(struct TreeNode* root)
 {
     int maxWithRoot = ;
     int maxWithoutRoot = ;

     traverse( root, &maxWithRoot, &maxWithoutRoot );

     return MAX(maxWithRoot, maxWithoutRoot);
 }