由前序遍历和中序遍历构造二叉树

 

 

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/**  * Definition for a binary tree node.  * type TreeNode struct {  *     Val int  *     Left *TreeNode  *     Right *TreeNode  * }  */ func buildTree(preorder []int, inorder []int) *TreeNode {     return build(preorder,0,len(preorder)-1,inorder,0,len(inorder)-1) }   func build(preorder []int,s1,e1 int,inorder []int,s2,e2 int)*TreeNode{     if s1>e1{         return nil     }     // root 节点对应的值就是前序遍历数组的第⼀个元素     rootVal:=preorder[s1]     // rootVal 在中序遍历数组中的索引     inIdx:=s2     for i:=s2;i<e2;i++{         if inorder[i]==rootVal{             inIdx=i             break         }     }     root:=&TreeNode{}     root.Val=rootVal     // 递归构造左右⼦树     leftSize:=inIdx-s2//左子树节点个数     root.Left=build(preorder,s1+1,s1+leftSize,inorder,s2,inIdx-1)     root.Right=build(preorder,s1+leftSize+1,e1,inorder,inIdx+1,e2)     return root }