Formatted question description: https://leetcode.ca/all/783.html

783. Minimum Distance Between BST Nodes (Medium)

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:

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

Solution 1.

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int minDiffInBST(TreeNode root) {
List<Integer> inorderTraversal = inorderTraversal(root);
int minDiff = Integer.MAX_VALUE;
int size = inorderTraversal.size();
for (int i = 1; i < size; i++) {
int diff = inorderTraversal.get(i) - inorderTraversal.get(i - 1);
minDiff = Math.min(minDiff, diff);
}
return minDiff;
}

public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> inorderTraversal = new ArrayList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode node = root;
while (!stack.isEmpty() || node != null) {
while (node != null) {
stack.push(node);
node = node.left;
}
TreeNode visitNode = stack.pop();
node = visitNode.right;
}
return inorderTraversal;
}
}

############

/**
* 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 {
private int ans;
private int prev;
private int inf = Integer.MAX_VALUE;

public int minDiffInBST(TreeNode root) {
ans = inf;
prev = inf;
dfs(root);
return ans;
}

private void dfs(TreeNode root) {
if (root == null) {
return;
}
dfs(root.left);
ans = Math.min(ans, Math.abs(root.val - prev));
prev = root.val;
dfs(root.right);
}
}

• // OJ: https://leetcode.com/problems/minimum-distance-between-bst-nodes/
// Time: O(N)
// Space: O(log(N))
class Solution {
private:
int ans = INT_MAX;
TreeNode *prev = NULL;
void inorder(TreeNode *root) {
if (!root) return;
inorder(root->left);
if (prev) ans = min(ans, root->val - prev->val);
prev = root;
inorder(root->right);
}
public:
int minDiffInBST(TreeNode* root) {
inorder(root);
return ans;
}
};

• # 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 minDiffInBST(self, root: Optional[TreeNode]) -> int:
def dfs(root):
if root is None:
return
dfs(root.left)
nonlocal ans, prev
ans = min(ans, abs(prev - root.val))
prev = root.val
dfs(root.right)

ans = prev = inf
dfs(root)
return ans

############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
def minDiffInBST(self, root):
"""
:type root: TreeNode
:rtype: int
"""
vals = []
def inOrder(root):
if not root:
return
inOrder(root.left)
vals.append(root.val)
inOrder(root.right)
inOrder(root)
return min([vals[i + 1] - vals[i] for i in xrange(len(vals) - 1)])

• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func minDiffInBST(root *TreeNode) int {
inf := 0x3f3f3f3f
ans, prev := inf, inf
var dfs func(*TreeNode)
dfs = func(root *TreeNode) {
if root == nil {
return
}
dfs(root.Left)
ans = min(ans, abs(prev-root.Val))
prev = root.Val
dfs(root.Right)
}
dfs(root)
return ans
}

func min(a, b int) int {
if a < b {
return a
}
return b
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}