106. 从中序与后序遍历序列构造二叉树

原题

https://leetcode.cn/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/

思路

递归

复杂度

时间：O(n)
空间：O(n)

Python代码

# 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
class Solution:
    def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
        if not postorder:
            return None
        root = TreeNode(postorder[-1])
        # 根节点在中序遍历数组的索引
        idx = inorder.index(postorder[-1])
        root.left = self.buildTree(inorder[:idx], postorder[:idx])
        root.right = self.buildTree(inorder[idx + 1 :], postorder[idx:-1])
        return root

Go代码

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func buildTree(inorder []int, postorder []int) *TreeNode {
	if len(postorder) == 0 {
		return nil
	}
	root := &TreeNode{Val: postorder[len(postorder)-1]}
	// 根节点在中序遍历数组的索引
	idx := slices.Index(inorder, postorder[len(postorder)-1])
	root.Left = buildTree(inorder[:idx], postorder[:idx])
	root.Right = buildTree(inorder[idx+1:], postorder[idx:len(postorder)-1])
	return root
}