LWC 71: 783. Minimum Distance Between BST Nodes

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