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73 lines
1.7 KiB
Markdown
Executable File
73 lines
1.7 KiB
Markdown
Executable File
# [105. Construct Binary Tree from Preorder and Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)
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## 题目
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Given preorder and inorder traversal of a tree, construct the binary tree.
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**Note**:You may assume that duplicates do not exist in the tree.
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For example, given
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preorder = [3,9,20,15,7]
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inorder = [9,3,15,20,7]
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Return the following binary tree:
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3
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/ \
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9 20
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/ \
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15 7
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## 题目大意
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根据一棵树的前序遍历与中序遍历构造二叉树。
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注意:
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你可以假设树中没有重复的元素。
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## 解题思路
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- 给出 2 个数组，根据 preorder 和 inorder 数组构造一颗树。
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- 利用递归思想，从 preorder 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。
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## 代码
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```go
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package leetcode
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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func buildTree(preorder []int, inorder []int) *TreeNode {
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inPos := make(map[int]int)
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for i := 0; i < len(inorder); i++ {
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inPos[inorder[i]] = i
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}
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return buildPreIn2TreeDFS(preorder, 0, len(preorder)-1, 0, inPos)
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}
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func buildPreIn2TreeDFS(pre []int, preStart int, preEnd int, inStart int, inPos map[int]int) *TreeNode {
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if preStart > preEnd {
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return nil
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}
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root := &TreeNode{Val: pre[preStart]}
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rootIdx := inPos[pre[preStart]]
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leftLen := rootIdx - inStart
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root.Left = buildPreIn2TreeDFS(pre, preStart+1, preStart+leftLen, inStart, inPos)
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root.Right = buildPreIn2TreeDFS(pre, preStart+leftLen+1, preEnd, rootIdx+1, inPos)
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return root
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}
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``` |