@@ -388,6 +388,90 @@ class Solution:
388388
389389### Go
390390
391+ 暴力递归
392+ 393+ ``` go
394+ /* *
395+ * Definition for a binary tree node.
396+ * type TreeNode struct {
397+ * Val int
398+ * Left *TreeNode
399+ * Right *TreeNode
400+ * }
401+ */
402+ func rob (root *TreeNode ) int {
403+ if root == nil {
404+ return 0
405+ }
406+ if root.Left == nil && root.Right == nil {
407+ return root.Val
408+ }
409+ // 偷父节点
410+ val1 := root.Val
411+ if root.Left != nil {
412+ val1 += rob (root.Left .Left ) + rob (root.Left .Right ) // 跳过root->left,相当于不考虑左孩子了
413+ }
414+ if root.Right != nil {
415+ val1 += rob (root.Right .Left ) + rob (root.Right .Right ) // 跳过root->right,相当于不考虑右孩子了
416+ }
417+ // 不偷父节点
418+ val2 := rob (root.Left ) + rob (root.Right ) // 考虑root的左右孩子
419+ return max (val1, val2)
420+ }
421+ 422+ func max (x , y int ) int {
423+ if x > y {
424+ return x
425+ }
426+ return y
427+ }
428+ ```
429+ 430+ 记忆化递推
431+ 432+ ``` go
433+ /* *
434+ * Definition for a binary tree node.
435+ * type TreeNode struct {
436+ * Val int
437+ * Left *TreeNode
438+ * Right *TreeNode
439+ * }
440+ */
441+ var umap = make (map [*TreeNode]int )
442+ 443+ func rob (root *TreeNode ) int {
444+ if root == nil {
445+ return 0
446+ }
447+ if root.Left == nil && root.Right == nil {
448+ return root.Val
449+ }
450+ if val , ok := umap[root]; ok {
451+ return val // 如果umap里已经有记录则直接返回
452+ }
453+ // 偷父节点
454+ val1 := root.Val
455+ if root.Left != nil {
456+ val1 += rob (root.Left .Left ) + rob (root.Left .Right ) // 跳过root->left,相当于不考虑左孩子了
457+ }
458+ if root.Right != nil {
459+ val1 += rob (root.Right .Left ) + rob (root.Right .Right ) // 跳过root->right,相当于不考虑右孩子了
460+ }
461+ // 不偷父节点
462+ val2 := rob (root.Left ) + rob (root.Right ) // 考虑root的左右孩子
463+ umap[root] = max (val1, val2) // umap记录一下结果
464+ return max (val1, val2)
465+ }
466+ 467+ func max (x , y int ) int {
468+ if x > y {
469+ return x
470+ }
471+ return y
472+ }
473+ ```
474+ 391475动态规划
392476
393477``` go
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