Question

1008. 前序遍历构造二叉搜索树

English Version

题目描述

给定一个整数数组，它表示BST(即 二叉搜索树 )的 先序遍历 ，构造树并返回其根。

保证 对于给定的测试用例，总是有可能找到具有给定需求的二叉搜索树。

二叉搜索树 是一棵二叉树，其中每个节点， Node.left 的任何后代的值 严格小于 Node.val , Node.right 的任何后代的值 严格大于 Node.val。

二叉树的 前序遍历 首先显示节点的值，然后遍历Node.left，最后遍历Node.right。

 

示例 1：

输入：preorder = [8,5,1,7,10,12]
输出：[8,5,10,1,7,null,12]

示例 2:

输入: preorder = [1,3]
输出: [1,null,3]

 

提示：

• 1 <= preorder.length <= 100
• 1 <= preorder[i] <= 10^8
• preorder 中的值 互不相同

 

解法

方法一

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def bstFromPreorder(self, preorder: List[int]) -> Optional[TreeNode]:
    def dfs(preorder):
      if not preorder:
        return None
      root = TreeNode(preorder[0])
      left, right = 1, len(preorder)
      while left < right:
        mid = (left + right) >> 1
        if preorder[mid] > preorder[0]:
          right = mid
        else:
          left = mid + 1
      root.left = dfs(preorder[1:left])
      root.right = dfs(preorder[left:])
      return root

    return dfs(preorder)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {

  public TreeNode bstFromPreorder(int[] preorder) {
    return dfs(preorder, 0, preorder.length - 1);
  }

  private TreeNode dfs(int[] preorder, int i, int j) {
    if (i > j || i >= preorder.length) {
      return null;
    }
    TreeNode root = new TreeNode(preorder[i]);
    int left = i + 1, right = j + 1;
    while (left < right) {
      int mid = (left + right) >> 1;
      if (preorder[mid] > preorder[i]) {
        right = mid;
      } else {
        left = mid + 1;
      }
    }
    root.left = dfs(preorder, i + 1, left - 1);
    root.right = dfs(preorder, left, j);
    return root;
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  TreeNode* bstFromPreorder(vector<int>& preorder) {
    return dfs(preorder, 0, preorder.size() - 1);
  }

  TreeNode* dfs(vector<int>& preorder, int i, int j) {
    if (i > j || i >= preorder.size()) return nullptr;
    TreeNode* root = new TreeNode(preorder[i]);
    int left = i + 1, right = j + 1;
    while (left < right) {
      int mid = (left + right) >> 1;
      if (preorder[mid] > preorder[i])
        right = mid;
      else
        left = mid + 1;
    }
    root->left = dfs(preorder, i + 1, left - 1);
    root->right = dfs(preorder, left, j);
    return root;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func bstFromPreorder(preorder []int) *TreeNode {
  var dfs func(i, j int) *TreeNode
  dfs = func(i, j int) *TreeNode {
    if i > j || i >= len(preorder) {
      return nil
    }
    root := &TreeNode{Val: preorder[i]}
    left, right := i+1, len(preorder)
    for left < right {
      mid := (left + right) >> 1
      if preorder[mid] > preorder[i] {
        right = mid
      } else {
        left = mid + 1
      }
    }
    root.Left = dfs(i+1, left-1)
    root.Right = dfs(left, j)
    return root
  }
  return dfs(0, len(preorder)-1)
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function bstFromPreorder(preorder: number[]): TreeNode | null {
  const n = preorder.length;
  const next = new Array(n);
  const stack = [];
  for (let i = n - 1; i >= 0; i--) {
    while (stack.length !== 0 && preorder[stack[stack.length - 1]] < preorder[i]) {
      stack.pop();
    }
    next[i] = stack[stack.length - 1] ?? n;
    stack.push(i);
  }

  const dfs = (left: number, right: number) => {
    if (left >= right) {
      return null;
    }
    return new TreeNode(preorder[left], dfs(left + 1, next[left]), dfs(next[left], right));
  };
  return dfs(0, n);
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//   TreeNode {
//     val,
//     left: None,
//     right: None
//   }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
  fn dfs(
    preorder: &Vec<i32>,
    next: &Vec<usize>,
    left: usize,
    right: usize
  ) -> Option<Rc<RefCell<TreeNode>>> {
    if left >= right {
      return None;
    }
    Some(
      Rc::new(
        RefCell::new(TreeNode {
          val: preorder[left],
          left: Self::dfs(preorder, next, left + 1, next[left]),
          right: Self::dfs(preorder, next, next[left], right),
        })
      )
    )
  }

  pub fn bst_from_preorder(preorder: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
    let n = preorder.len();
    let mut stack = Vec::new();
    let mut next = vec![n; n];
    for i in (0..n).rev() {
      while !stack.is_empty() && preorder[*stack.last().unwrap()] < preorder[i] {
        stack.pop();
      }
      if !stack.is_empty() {
        next[i] = *stack.last().unwrap();
      }
      stack.push(i);
    }
    Self::dfs(&preorder, &next, 0, n)
  }
}