[LeetCode] 783. Minimum Distance Between BST Nodes

轉自LeetCode

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>

想法如下

• 用 inorder 的方式歷遍 BST，因為是 BST，所以結果會是 sorted 的

• 在歷遍的同時，紀錄前一個數字，並計算當前數字減掉前一個數字，是不是最小值

• 只要計算 “當前數字減掉前一個數字” 是因為歷遍結果是 sorted 的，加上是要找最小值，所以只要檢查相鄰兩個數就可以了

code 如下

Java(參考解答)

	/**
	* Definition for a binary tree node.
	* public class TreeNode {
	* int val;
	* TreeNode left;
	* TreeNode right;
	* TreeNode(int x) { val = x; }
	* }
	*/
	class Solution {
	Integer mPrv = null;
	Integer mMin = Integer.MAX_VALUE;
	public int minDiffInBST(TreeNode root) {
	inorder(root);
	return mMin;
	}
	private void inorder(TreeNode root) {
	if(root == null) {
	return;
	}
	inorder(root.left);
	if(mPrv != null) {
	mMin = Math.min(mMin, root.val - mPrv);
	}
	mPrv = root.val;
	inorder(root.right);
	}
	}

view raw minDiffInBST.java hosted with ❤ by GitHub