由前序遍历和中序遍历构造二叉树

 

 

/**
 * 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
}