[LeetCode] 530. Minimum Absolute Difference in BST

轉自LeetCode

Given a binary search tree with non-negative values, find the minimum absolute difference between values of any two nodes.

Example:

Input:

   1
    \
     3
    /
   2

Output:
1

Explanation:
The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).

Note: There are at least two nodes in this BST.

<Solution>

想法如下

• 因為是BST，所以用 inorder traversal 會得到一個由小到大的數列。因此，在歷遍的時候，檢查當下的值減掉前面的值，並記錄最小的差距，歷遍結束後就會得到答案。這是因為數列是排序的，最小值一定是兩個相鄰的數相減來得到的

code 如下

C++(使用遞迴)

	/**
	* Definition for a binary tree node.
	* struct TreeNode {
	* int val;
	* TreeNode *left;
	* TreeNode *right;
	* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	* };
	*/
	class Solution {
	public:
	int ans = INT_MAX;
	int prv = -1;
	int getMinimumDifference(TreeNode* root) {
	return inorder(root);
	}
	int inorder(TreeNode *node) {
	if(node == NULL) {
	return ans;
	}
	inorder(node->left);
	if(prv != -1) {
	ans = min(ans, node->val - prv);
	}
	prv = node->val;
	inorder(node->right);
	return ans;
	}
	};

view raw MinAbsInBST.cpp hosted with ❤ by GitHub

Java(使用 stack)

	/**
	* Definition for a binary tree node.
	* public class TreeNode {
	* int val;
	* TreeNode left;
	* TreeNode right;
	* TreeNode(int x) { val = x; }
	* }
	*/
	class Solution {
	public int getMinimumDifference(TreeNode root) {
	Stack<TreeNode> stack = new Stack<>();
	int ans = Integer.MAX_VALUE;
	while(root != null) {
	stack.push(root);
	root = root.left;
	}
	TreeNode node;
	int prv = -1;
	while(!stack.empty()) {
	node = stack.pop();
	if(prv != -1) {
	ans = Math.min(ans, node.val - prv);
	}
	prv = node.val;
	if(node.right != null) {
	node = node.right;
	while(node != null) {
	stack.push(node);
	node = node.left;
	}
	}
	}
	return ans;
	}
	}

view raw MinAbsInBST.java hosted with ❤ by GitHub