[Golang][Leetcode][BinaryTree]刷題系列-106-Construct Binary Tree from Inorder and Postorder Traversal

2021-09-04 269 words 2 minutes

/leetcode106_binarytree/golang_leetcode.png

Contents

106. Construct Binary Tree from Inorder and Postorder Traversal

Level : Medium

原題連結 : Click

題目 :

Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.

Example :

Note

Example 1:

Input: inorder = [9,3,15,20,7], postorder = [9,15,7,20,3] Output: [3,9,20,null,null,15,7]

Example 2:

Input: inorder = [-1], postorder = [-1] Output: [-1]

解題思路 :

這個也是一個非常經典的題目! 題目給你兩組數組，分別是inorder排序與postorder排序，需要用兩個數組的特性去建構出linked node版本的binary tree
inorder的特性是left -> mid -> right
postorder的特性是 left -> right -> mid
由於以上兩種特性我可以做以下操作:
1. 先取得postorder數組的最後一個element，他必定是mid(root)
2. 再用該element去inorder中尋找，這樣就能依照inorder的規則去判斷左邊是Left tree 右邊是 right tree，將其與選出的root連接
3. 重複以上動作

Recursive解法

time complexity: O(n^2) , space complexity: O(n)

Runtime: 4 ms, faster than 90.98% of Go online submissions for Construct Binary Tree from Inorder and Postorder Traversal.

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/**
 * Definition for a binary tree node.
 * 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
    
    midVal := postorder[len(postorder)-1]
    
    
    midIndex := findIndex(midVal,inorder)
    
    node := &TreeNode {
        Val: midVal,
        Left: buildTree(inorder[:midIndex], postorder[:midIndex]),
        Right: buildTree(inorder[midIndex+1:], postorder[midIndex:len(postorder)-1]),
    }

    return node
}

func findIndex(tar int, arr []int) (index int) {
    
    for i, v := range arr {
        if v == tar {
            return i
        }
    }
    
    return -1
}

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