1028. Recover a Tree From Preorder Traversal

1028. Recover a Tree From Preorder Traversal #

题目 #

We run a preorder depth first search on the root of a binary tree.

At each node in this traversal, we output D dashes (where D is the depth of this node), then we output the value of this node. (If the depth of a node is D, the depth of its immediate child is D+1. The depth of the root node is 0.)

If a node has only one child, that child is guaranteed to be the left child.

Given the output S of this traversal, recover the tree and return its root.

Example 1:

https://assets.leetcode.com/uploads/2019/04/08/recover-a-tree-from-preorder-traversal.png

Input: "1-2--3--4-5--6--7"
Output: [1,2,5,3,4,6,7]

Example 2:

https://assets.leetcode.com/uploads/2019/04/11/screen-shot-2019-04-10-at-114101-pm.png

Input: "1-2--3---4-5--6---7"
Output: [1,2,5,3,null,6,null,4,null,7]

Example 3:

https://assets.leetcode.com/uploads/2019/04/11/screen-shot-2019-04-10-at-114955-pm.png

Input: "1-401--349---90--88"
Output: [1,401,null,349,88,90]

Note:

  • The number of nodes in the original tree is between 1 and 1000.
  • Each node will have a value between 1 and 10^9.

题目大意 #

我们从二叉树的根节点 root 开始进行深度优先搜索。

在遍历中的每个节点处,我们输出 D 条短划线(其中 D 是该节点的深度),然后输出该节点的值。(如果节点的深度为 D,则其直接子节点的深度为 D + 1。根节点的深度为 0)。如果节点只有一个子节点,那么保证该子节点为左子节点。给出遍历输出 S,还原树并返回其根节点 root。

提示:

  • 原始树中的节点数介于 1 和 1000 之间。
  • 每个节点的值介于 1 和 10 ^ 9 之间。

解题思路 #

  • 给出一个字符串,字符串是一个树的先根遍历的结果,其中破折号的个数代表层数。请根据这个字符串生成对应的树。
  • 这一题解题思路比较明确,用 DFS 就可以解题。边深搜字符串,边根据破折号的个数判断当前节点是否属于本层。如果不属于本层,回溯到之前的根节点,添加叶子节点以后再继续深搜。需要注意的是每次深搜时,扫描字符串的 index 需要一直保留,回溯也需要用到这个 index。

代码 #


package leetcode

import (
	"strconv"
)

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func recoverFromPreorder(S string) *TreeNode {
	if len(S) == 0 {
		return &TreeNode{}
	}
	root, index, level := &TreeNode{}, 0, 0
	cur := root
	dfsBuildPreorderTree(S, &index, &level, cur)
	return root.Right
}

func dfsBuildPreorderTree(S string, index, level *int, cur *TreeNode) (newIndex *int) {
	if *index == len(S) {
		return index
	}
	if *index == 0 && *level == 0 {
		i := 0
		for i = *index; i < len(S); i++ {
			if !isDigital(S[i]) {
				break
			}
		}
		num, _ := strconv.Atoi(S[*index:i])
		tmp := &TreeNode{Val: num, Left: nil, Right: nil}
		cur.Right = tmp
		nLevel := *level + 1
		index = dfsBuildPreorderTree(S, &i, &nLevel, tmp)
		index = dfsBuildPreorderTree(S, index, &nLevel, tmp)
	}
	i := 0
	for i = *index; i < len(S); i++ {
		if isDigital(S[i]) {
			break
		}
	}
	if *level == i-*index {
		j := 0
		for j = i; j < len(S); j++ {
			if !isDigital(S[j]) {
				break
			}
		}
		num, _ := strconv.Atoi(S[i:j])
		tmp := &TreeNode{Val: num, Left: nil, Right: nil}
		if cur.Left == nil {
			cur.Left = tmp
			nLevel := *level + 1
			index = dfsBuildPreorderTree(S, &j, &nLevel, tmp)
			index = dfsBuildPreorderTree(S, index, level, cur)
		} else if cur.Right == nil {
			cur.Right = tmp
			nLevel := *level + 1
			index = dfsBuildPreorderTree(S, &j, &nLevel, tmp)
			index = dfsBuildPreorderTree(S, index, level, cur)
		}
	}
	return index
}


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