JavaScript刷LeetCode -- 971. Flip Binary Tree To Match Preorder Traversal [Medium]

一、题目

  Given a binary tree with N nodes, each node has a different value from {1, …, N}.

  A node in this binary tree can be flipped by swapping the left child and the right child of that node.

  Consider the sequence of N values reported by a preorder traversal starting from the root. Call such a sequence of N values the voyage of the tree.

  (Recall that a preorder traversal of a node means we report the current node’s value, then preorder-traverse the left child, then preorder-traverse the right child.)

  Our goal is to flip the least number of nodes in the tree so that the voyage of the tree matches the voyage we are given.

  If we can do so, then return a list of the values of all nodes flipped. You may return the answer in any order.

  If we cannot do so, then return the list [-1].

二、题目大意

  给定一棵二叉树，能够通过翻转节点的左右子节点得到一个前序序列voyage，返回翻转节点的集合，否则返回[-1]。

三、解题思路

  这道题实际上就是对于前序遍历二叉树的扩展，在前序遍历的过程中与voyage数组中的节点进行比较。

四、代码实现

const flipMatchVoyage = (root, voyage) => {
  let ans = []
  // 记录访问到第几个节点
  let position = 0

  help(root, voyage, position)

  return ans

  function help (root, voyage) {
    if (!root) {
      return
    }
    if (root.val !== voyage[position++]) {
      ans = [-1]
      return
    }

    if (root.left && root.left.val !== voyage[position]) {
      // 更改遍历顺序达到交换左右子树的效果
      ans.push(root.val)
      help(root.right, voyage)
      help(root.left, voyage)
    } else {
      help(root.left, voyage)
      help(root.right, voyage)
    }
  }
}

  如果本文对您有帮助，欢迎关注微信公众号，为您推送更多内容，ε=ε=ε=┏(゜ロ゜;)┛。