剑指 Offer 07. 重建二叉树 (golang版)

• 思路：
  1.前序遍历的第一个节点是二叉树的根节点
  2.中序遍历中，根节点左边是左子树，右边是右子树
  3.理清思绪递归实现即可

func buildTree(preorder []int, inorder []int) *TreeNode {
	//结束条件，最下面的叶子节点时结束
	if len(preorder) == 0 {
		return nil
	}
	//根据前序遍历的性质在中序遍历中找到根节点的位置
	var i = 0
	for i < len(preorder) && preorder[0] != inorder[i] {
		i++
	}
	//递归左子树
	l := buildTree(preorder[1:i+1], inorder[:i])
	//递归右子树
	r := buildTree(preorder[i+1:], inorder[i+1:])

	return &TreeNode{
		Val: preorder[0],
		Left: l,
		Right: r,
	}
}

• 测试代码：

package main

import "fmt"

type TreeNode struct {
	Val int
	Left *TreeNode
	Right *TreeNode
}

func main() {
	//var preorder = []int{3,9,20,15,7}
	//var inorder = []int {9,3,15,20,7}
	var preorder = []int{1,2,4,8,5,9,3,6,7,10}
	var inorder =  []int{4,8,2,9,5,1,6,3,7,10}
	root := buildTree(preorder,inorder)
	//前序遍历
	//printVLRTree(root)
	//中序遍历
	//printLDRTree(root)
	//后序遍历
	printLRDTree(root)
}



func buildTree(preorder []int, inorder []int) *TreeNode {
	if len(preorder) == 0 {
		return nil
	}

	var i = 0
	for i < len(preorder) && preorder[0] != inorder[i] {
		i++
	}
	//fmt.Println(preorder,inorder,i)
	l := buildTree(preorder[1:i+1], inorder[:i])
	r := buildTree(preorder[i+1:], inorder[i+1:])

	return &TreeNode{
		Val: preorder[0],
		Left: l,
		Right: r,
	}
}

//前序遍历
func printVLRTree(root *TreeNode){
	if root == nil {
		return
	}
	fmt.Println(root.Val)
	printVLRTree(root.Left)
	printVLRTree(root.Right)
}
//中序遍历
func printLDRTree(root *TreeNode){
	if root == nil {
		return
	}
	printLDRTree(root.Left)
	fmt.Println(root.Val)
	printLDRTree(root.Right)
}
//后序遍历
func printLRDTree(root *TreeNode){
	if root == nil {
		return
	}
	printLRDTree(root.Left)
	printLRDTree(root.Right)
	fmt.Println(root.Val)
}