106. Construct Binary Tree from Inorder and Postorder Traversal #

Problem #

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:

Summary #

Construct a binary tree from the inorder traversal and postorder traversal of a tree.

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

Solution Idea #

Given 2 arrays, construct a tree from the inorder and postorder arrays.
Use recursion. The root node can be obtained from postorder, and the left and right subtrees can be obtained from inorder. When only one node remains, it is the root node. Keep recursing until all trees have been generated.

Code #


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
 * }
 */

// Solution 1, directly pass in the required slice range as input, which can avoid allocating memory for the corresponding inorder indices; memory usage (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
}

// Solution 2
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
}