0105. Construct Binary Tree From Preorder and Inorder Traversal

105. Construct Binary Tree from Preorder and Inorder Traversal #

题目 #

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

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

For example, given

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

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

题目大意 #

根据一棵树的前序遍历与中序遍历构造二叉树。

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 preorder 和 inorder 数组构造一颗树。
  • 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。

代码 #


package leetcode

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func buildTree(preorder []int, inorder []int) *TreeNode {
	inPos := make(map[int]int)
	for i := 0; i < len(inorder); i++ {
		inPos[inorder[i]] = i
	}
	return buildPreIn2TreeDFS(preorder, 0, len(preorder)-1, 0, inPos)
}

func buildPreIn2TreeDFS(pre []int, preStart int, preEnd int, inStart int, inPos map[int]int) *TreeNode {
	if preStart > preEnd {
		return nil
	}
	root := &TreeNode{Val: pre[preStart]}
	rootIdx := inPos[pre[preStart]]
	leftLen := rootIdx - inStart
	root.Left = buildPreIn2TreeDFS(pre, preStart+1, preStart+leftLen, inStart, inPos)
	root.Right = buildPreIn2TreeDFS(pre, preStart+leftLen+1, preEnd, rootIdx+1, inPos)
	return root
}


⬅️上一页

下一页➡️

Calendar Sep 6, 2020
Edit Edit this page
本站总访问量:  次 您是本站第  位访问者