leeocode地址：从中序与后序遍历序列构造二叉树
给定两个整数数组 inorder 和 postorder ，其中 inorder 是二叉树的中序遍历， postorder 是同一棵树的后序遍历，请你构造并返回这颗 二叉树 。
示例 1:
输入：inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
输出：[3,9,20,null,null,15,7]
示例 2:
输入：inorder = [-1], postorder = [-1]
输出：[-1]
提示:
1 <= inorder.length <= 3000
postorder.length == inorder.length
-3000 <= inorder[i], postorder[i] <= 3000
inorder 和 postorder 都由 不同 的值组成
postorder 中每一个值都在 inorder 中
inorder 保证是树的中序遍历
postorder 保证是树的后序遍历
实现思路
中序遍历（Inorder）：左子树 -> 根节点 -> 右子树
后序遍历（Postorder）：左子树 -> 右子树 -> 根节点
通过给定的中序遍历和后序遍历数组，我们可以确定二叉树的根节点以及左右子树的范围。具体步骤如下：
步骤1：后序遍历的最后一个元素是根节点的值。
步骤2：在中序遍历中找到根节点的位置，其左侧为左子树的中序遍历，右侧为右子树的中序遍历。
步骤3：根据步骤2中左右子树的大小，可以在后序遍历中确定左子树和右子树的后序遍历。
递归地应用以上步骤，即可构造整棵二叉树。
代码实现
# Definition for a binary tree node.
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def buildTree(inorder, postorder):
if not inorder or not postorder:
return None
root_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 root
def 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实现
package main
import "fmt"
type TreeNode struct {
Val int
Left *TreeNode
Right *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 []int
var 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() {
// Example
inorder := []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 traversal
fmt.Println("Inorder traversal of constructed tree:", inorderTraversal(root))
}kotlin实现
class TreeNode(var `val`: Int) {
var left: TreeNode? = null
var 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) return
inorder(node.left)
result.add(node.`val`)
inorder(node.right)
}
inorder(root)
return result
}
fun main() {
// Example
val 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 traversal
println("Inorder traversal of constructed tree: ${inorderTraversal(root)}")
}