337. 打家劫舍 III

暴力递归（超时）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int rob(TreeNode root) {
        if(root == null) return 0;
        if(root.left == null && root.right == null) return root.val;
        //偷父节点
        int val1 = root.val;
        if(root.left != null) {
            val1 += rob(root.left.left) + rob(root.left.right);
        }
        if(root.right != null) {
            val1 += rob(root.right.left) + rob(root.right.right);
        }
        //不偷父节点
        int val2 = rob(root.left) + rob(root.right);
        return Math.max(val1,val2);
    }
}

记忆化

class Solution {
    Map<TreeNode, Integer> map = new HashMap<>();
    public int rob(TreeNode root) {
        if(root == null) return 0;
        if(root.left == null && root.right == null) return root.val;
        if(map.containsKey(root) ) {
            return map.get(root);
        }
        //偷父节点
        int val1 = root.val;
        if(root.left != null) {
            val1 += rob(root.left.left) + rob(root.left.right);
        }
        if(root.right != null) {
            val1 += rob(root.right.left) + rob(root.right.right);
        }
        //不偷父节点
        int val2 = rob(root.left) + rob(root.right);
        map.put(root,Math.max(val1,val2) );
        return Math.max(val1,val2);
    }
}

动态规划

class Solution {
    public int rob(TreeNode root) {
        //dp[0]表示不偷该节点
        //dp[1]表示偷该节点
        int [] dp = myRob(root);
        return Math.max(dp[0],dp[1]);
    }
    public int [] myRob(TreeNode root ) {
        int [] dp = new int [2];
        if(null == root ) {
            return dp; //初始化为{0，0}
        }
        int [] left = myRob(root.left);
        int [] right = myRob(root.right );
        //不偷当前节点，取左右子节点最大值 加和
        dp[0] = Math.max(left[0],left[1]) + Math.max(right[0],right[1]) ;
        dp[1] = root.val + left[0] + right[0];
        return dp;
    }
}