783. Minimum Distance Between BST Nodes

#### QUESTION:

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.

#### EXPLANATION:

1.因为求两个数字之间的最小距离，所以可以取出所有的值，然后再进行比较。

2.在遍历的过程中就进行比较。

#### SOLUTION:

``````class Solution {
Stack<Integer> minDiffInBSTStack = new Stack<>();
public int minDiffInBST(TreeNode root) {
minDiffInBSTHelper(root);
int[] tmp = new int[minDiffInBSTStack.size()];
int index = 0;
while (!minDiffInBSTStack.isEmpty()){
tmp[index] = minDiffInBSTStack.pop();
index++;
}
Arrays.sort(tmp);
int result = Integer.MAX_VALUE;
for(int i = 0;i<tmp.length-1;i++){
result = Math.min(result,Math.abs(tmp[i+1]-tmp[i]));
}
return result;
}
public void minDiffInBSTHelper(TreeNode root){
if(root==null)return;
minDiffInBSTStack.push(root.val);
if(root.left!=null) minDiffInBSTHelper(root.left);
if(root.right!=null) minDiffInBSTHelper(root.right);
}
}
``````
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