106. Construct Binary Tree from Inorder and Postorder Traversal #

题目 #

Given inorder and postorder traversal of a tree, construct the binary tree.

Note: You may assume that duplicates do not exist in the tree.

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

题目大意 #

根据一棵树的中序遍历与后序遍历构造二叉树。

注意: 你可以假设树中没有重复的元素。

解题思路 #

给出 2 个数组，根据 inorder 和 postorder 数组构造一颗树。
利用递归思想，从 postorder 可以得到根节点，从 inorder 中得到左子树和右子树。只剩一个节点的时候即为根节点。不断的递归直到所有的树都生成完成。

代码 #


package leetcode

import (
	"github.com/halfrost/leetcode-go/structures"
)

// TreeNode define
type TreeNode = structures.TreeNode

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */

// 解法一, 直接传入需要的 slice 范围作为输入, 可以避免申请对应 inorder 索引的内存, 内存使用(leetcode test case) 4.7MB -> 4.3MB.
func buildTree(inorder []int, postorder []int) *TreeNode {
	postorderLen := len(postorder)
	if len(inorder) == 0 {
		return nil
	}
	root := &TreeNode{Val: postorder[postorderLen-1]}
	postorder = postorder[:postorderLen-1]
	for pos, node := range inorder {
		if node == root.Val {
			root.Left = buildTree(inorder[:pos], postorder[:len(inorder[:pos])])
			root.Right = buildTree(inorder[pos+1:], postorder[len(inorder[:pos]):])
		}
	}
	return root
}

// 解法二
func buildTree1(inorder []int, postorder []int) *TreeNode {
	inPos := make(map[int]int)
	for i := 0; i < len(inorder); i++ {
		inPos[inorder[i]] = i
	}
	return buildInPos2TreeDFS(postorder, 0, len(postorder)-1, 0, inPos)
}

func buildInPos2TreeDFS(post []int, postStart int, postEnd int, inStart int, inPos map[int]int) *TreeNode {
	if postStart > postEnd {
		return nil
	}
	root := &TreeNode{Val: post[postEnd]}
	rootIdx := inPos[post[postEnd]]
	leftLen := rootIdx - inStart
	root.Left = buildInPos2TreeDFS(post, postStart, postStart+leftLen-1, inStart, inPos)
	root.Right = buildInPos2TreeDFS(post, postStart+leftLen, postEnd-1, rootIdx+1, inPos)
	return root
}

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