leeocode地址：從中序與后序遍歷序列構造二叉樹
給定兩個整數數組 inorder 和 postorder ，其中 inorder 是二叉樹的中序遍歷， postorder 是同一棵樹的后序遍歷，請你構造并返回這顆 二叉樹 。

示例 1:

輸入：inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
輸出：[3,9,20,null,null,15,7]
示例 2:

輸入：inorder = [-1], postorder = [-1]
輸出：[-1]

提示:
1 <= inorder.length <= 3000
postorder.length == inorder.length
-3000 <= inorder[i], postorder[i] <= 3000
inorder 和 postorder 都由 不同 的值組成
postorder 中每一個值都在 inorder 中
inorder 保證是樹的中序遍歷
postorder 保證是樹的后序遍歷

實現思路

中序遍歷（Inorder）：左子樹 -> 根節點 -> 右子樹
后序遍歷（Postorder）：左子樹 -> 右子樹 -> 根節點
通過給定的中序遍歷和后序遍歷數組，我們可以確定二叉樹的根節點以及左右子樹的范圍。具體步驟如下：
步驟1：后序遍歷的最后一個元素是根節點的值。
步驟2：在中序遍歷中找到根節點的位置，其左側為左子樹的中序遍歷，右側為右子樹的中序遍歷。
步驟3：根據步驟2中左右子樹的大小，可以在后序遍歷中確定左子樹和右子樹的后序遍歷。
遞歸地應用以上步驟，即可構造整棵二叉樹。

代碼實現

# Definition for a binary tree node.
class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef buildTree(inorder, postorder):if not inorder or not postorder:return Noneroot_val = postorder.pop()root = TreeNode(root_val)idx = inorder.index(root_val)root.right = buildTree(inorder[idx + 1:], postorder)root.left = buildTree(inorder[:idx], postorder)return rootdef inorderTraversal(root):if not root:return []return inorderTraversal(root.left) + [root.val] + inorderTraversal(root.right)# Example
inorder = [9, 3, 15, 20, 7]
postorder = [9, 15, 7, 20, 3]root = buildTree(inorder, postorder)# Verify the constructed tree by printing its inorder traversal
print("Inorder traversal of constructed tree:", inorderTraversal(root))

go實現

package mainimport "fmt"type TreeNode struct {Val   intLeft  *TreeNodeRight *TreeNode
}func buildTree(inorder []int, postorder []int) *TreeNode {if len(inorder) == 0 || len(postorder) == 0 {return nil}rootVal := postorder[len(postorder)-1]root := &TreeNode{Val: rootVal}idx := indexOf(inorder, rootVal)root.Left = buildTree(inorder[:idx], postorder[:idx])root.Right = buildTree(inorder[idx+1:], postorder[idx:len(postorder)-1])return root
}func indexOf(arr []int, val int) int {for i := range arr {if arr[i] == val {return i}}return -1
}func inorderTraversal(root *TreeNode) []int {var result []intvar inorder func(node *TreeNode)inorder = func(node *TreeNode) {if node == nil {return}inorder(node.Left)result = append(result, node.Val)inorder(node.Right)}inorder(root)return result
}func main() {// Exampleinorder := []int{9, 3, 15, 20, 7}postorder := []int{9, 15, 7, 20, 3}root := buildTree(inorder, postorder)// Verify the constructed tree by printing its inorder traversalfmt.Println("Inorder traversal of constructed tree:", inorderTraversal(root))
}

kotlin實現

class TreeNode(var `val`: Int) {var left: TreeNode? = nullvar right: TreeNode? = null
}fun buildTree(inorder: IntArray, postorder: IntArray): TreeNode? {if (inorder.isEmpty() || postorder.isEmpty()) {return null}val rootVal = postorder.last()val root = TreeNode(rootVal)val idx = inorder.indexOf(rootVal)root.left = buildTree(inorder.sliceArray(0 until idx), postorder.sliceArray(0 until idx))root.right = buildTree(inorder.sliceArray(idx + 1 until inorder.size), postorder.sliceArray(idx until postorder.size - 1))return root
}fun inorderTraversal(root: TreeNode?): List<Int> {val result = mutableListOf<Int>()fun inorder(node: TreeNode?) {if (node == null) returninorder(node.left)result.add(node.`val`)inorder(node.right)}inorder(root)return result
}fun main() {// Exampleval inorder = intArrayOf(9, 3, 15, 20, 7)val postorder = intArrayOf(9, 15, 7, 20, 3)val root = buildTree(inorder, postorder)// Verify the constructed tree by printing its inorder traversalprintln("Inorder traversal of constructed tree: ${inorderTraversal(root)}")
}