【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. 

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.

题目大意：找到二叉搜索树中任意两节点之差最小。

分析：由二叉搜索树的特性可知，每个节点的左子树 < 根节点 < 右节点 ，根据此特性我们知道，差值最小一定会出现在相邻的两个节点，也就是当前遍历到的节点与其左子树，右子树差值中的最小值。因此我们可以利用中序遍历求解。

代码：

class Solution {
    Integer prev= null;
    Integer res = Integer.MAX_VALUE;
    
    public int minDiffInBST(TreeNode root) {
        inOrder(root);
        return res;
    }
    
    
    private void inOrder(TreeNode root){
        if(root == null) return ; 
        inOrder(root.left);
        
        if(prev!= null){
            res = Math.min(res,root.val-prev);
        }
        prev = root.val;
        inOrder(root.right);
    }
}