[LeetCode] 783. Minimum Distance Between BST Nodes

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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.

题意：给一个二叉搜索树，找到其中两个节点的最小差值
二叉搜索树，左孩子小于根节点，右孩子大于根节点
所以，二叉搜索树 有着中序有序的特点
我们可以对其中序遍历，把数提出来，判断相邻两节点class Solution {

private List<Integer> list;
    private void DFS(TreeNode root) {
        if (root.left != null)
            DFS(root.left);
        list.add(root.val);
        if (root.right != null)
            DFS(root.right);
    }
    public int minDiffInBST(TreeNode root) {
        list = new ArrayList<>();
        int min = 3276800;
        DFS(root);
        for (int i = 0; i < list.size() - 1; i++) {
            int temp = list.get(i+1) - list.get(i);
            min = Math.min(temp, min);
        }
        return min;
    }
｝

当然这个算法显然有些多余，其一：额外开了O(n)的空间，其二多遍历了一遍
既然知道是中序有序的，那我们可以直接在遍历的时候就判断

class Solution {
    private int pre = -1;
    private int min = 3276800;
    public void DFS(TreeNode root) {
        if (root == null)
            return;
        DFS(root.left);
        if (pre != -1)
            min = Math.min(min, root.val - pre);
        pre = root.val;
        DFS(root.right);
    }
    
    public int minDiffInBST(TreeNode root) {
        DFS(root);
        return min;
    }
}