LWC 71: 783. Minimum Distance Between BST Nodes

Posted 2022-12-15 Demon的黑与白

LWC 71: 783. Minimum Distance Between BST Nodes

传送门：783. Minimum Distance Between BST Nodes

Problem:

Given a Binary Search Tree (BST) with the root node root, return the minimum difference between the values of any two different nodes in the tree.

Example:

Input: root = [4,2,6,1,3,null,null]
Output: 1
Explanation:
Note that root is a TreeNode object, not an array.

The given tree [4,2,6,1,3,null,null] is represented by the following diagram:

      4
    /   \\
  2      6
 / \\    
1   3  

while the minimum difference in this tree is 1, it occurs between node 1 and node 2, also between node 3 and node 2.

Note:

• The size of the BST will be between 2 and 100.
• The BST is always valid, each node’s value is an integer, and each node’s value is different.

思路：
BST + 中序，再求minimum difference

代码如下：

    public int minDiffInBST(TreeNode root) 
        vis = new ArrayList<Integer>();
        dfs(root);
        int min = 0x3f3f3f3f;
        for (int i = 1; i < vis.size(); ++i) 
            int diff = vis.get(i) - vis.get(i - 1);
            min = Math.min(min, diff);
        
        return min;
    

    List<Integer> vis;
    public void dfs(TreeNode root) 
        if (root == null) return;
        dfs(root.left);
        vis.add(root.val);
        dfs(root.right);
    

直接在中序的时候求出答案。

代码如下：

    public int minDiffInBST(TreeNode root) 
        min = 0x3f3f3f3f;
        prv = -1;
        solve(root);
        return min;
    

    int min = 0x3f3f3f3f;
    int prv = -1;
    void solve(TreeNode root) 
        if (root == null) return;
        solve(root.left);
        if (prv != -1) 
            min = Math.min(min, root.val - prv);
        
        prv = root.val;
        solve(root.right);
    

Python版本：

class Solution(object):
    def minDiffInBST(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """

        self.minv = float('inf')
        self.prv = -1

        def go(root):
            if not root: return
            go(root.left)
            if self.prv != -1: self.minv = min(self.minv, root.val - self.prv)
            self.prv = root.val
            go(root.right)

        go(root)
        return self.minv