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

根据二叉搜索树的性质，DFS 构建即可。

Python3

# 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)

Java

/**
 * 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;
    }
}

C++

/**
 * 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;
    }
};

Go

/**
 * 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)
}

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