Update 0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal.md

UPDATE a new answer for less memory cost & FIX method name typo.

website/content/ChapterFour/0100~0199/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal.md

@@ -48,7 +48,7 @@ package leetcode
  *     Right *TreeNode
  * }
  */
-func buildTree106(inorder []int, postorder []int) *TreeNode {
+func buildTree(inorder []int, postorder []int) *TreeNode {
 	inPos := make(map[int]int)
 	for i := 0; i < len(inorder); i++ {
 		inPos[inorder[i]] = i
@@ -68,6 +68,22 @@ func buildInPos2TreeDFS(post []int, postStart int, postEnd int, inStart int, inP
 	return root
 }
 
+// 解法 2, 直接传入需要的 slice 范围作为输入, 可以避免申请对应 inorder 索引的内存, 内存使用(leetcode test case) 4.7MB -> 4.3MB.
+func buildTree(inorder []int, postorder []int) *TreeNode {
+	postorderLen := len(postorder)
+    if len(inorder) == 0 { 
+    	return nil 
+    }
+    root := &TreeNode{Val:postorder[postorderLen-1]}
+    postorder = postorder[:postorderLen-1]
+    for pos, node := range inorder {
+    	if node == root.Val {
+    		root.Left  = buildTree(inorder[:pos], postorder[:len(inorder[:pos])])
+    		root.Right = buildTree(inorder[pos+1:], postorder[len(inorder[:pos]):])
+    	}
+    }
+    return root
+}
 ```