剑指 Offer 07. 重建二叉树
问题描述：
输入某二叉树的前序遍历和中序遍历的结果，请重建该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。
例如，给出
前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树：3
/
9 20
/
15 7限制：
0 <= 节点个数 <= 5000
/**
* Definition for a binary tree node.
* class TreeNode(var _value: Int) {
* var value: Int = _value
* var left: TreeNode = null
* var right: TreeNode = null
* }
*/
//注意算清两侧长度
object Solution {
def buildTree(preorder: Array[Int], inorder: Array[Int]): TreeNode = {
val root = build(preorder, inorder, 0, preorder.length-1, 0, inorder.length-1)
return root
}
def build(preorder: Array[Int], inorder: Array[Int], pl: Int, pr: Int, il: Int, ir:Int): TreeNode = {
if (pl > pr || il > ir) return null
val pivot = inorder.indexOf(preorder(pl))
val root = new TreeNode(preorder(pl))
root.left = build(preorder, inorder, pl+1, pl + (pivot - il) , il, pivot-1)
root.right = build(preorder, inorder, pl + (pivot - il) + 1, pr, pivot+1, ir)
return root
}
}
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func buildTree(preorder []int, inorder []int) *TreeNode {
if len(preorder) == 0 {return nil}
index := 0
for i := 0; i < len(inorder); i++ {
if inorder[i] == preorder[0] {
index = i
break
}
}
root := &TreeNode{Val: preorder[0]}
root.Left = buildTree(preorder[1:index+1], inorder[0:index])
root.Right = buildTree(preorder[index+1:], inorder[index+1:])
return root
}