Minimum Absolute Difference in BST

Question

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.

题解

题意为找任意两个节点间绝对值差的最小值,根据二叉搜索树的特性,中序遍历即得有序数组,找出相邻两数的最小差值即为解。

Java - Recursive

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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private int min = Integer.MAX_VALUE;
private TreeNode prev = null;

public int getMinimumDifference(TreeNode root) {
if (root == null) return min;

getMinimumDifference(root.left);
if (prev != null) {
min = Math.min(min, root.val - prev.val);
}
prev = root;
getMinimumDifference(root.right);
return min;
}
}

Java - Iterative

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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int getMinimumDifference(TreeNode root) {
int min = Integer.MAX_VALUE;
TreeNode prev = null;
Deque<TreeNode> stack = new ArrayDeque<TreeNode>();

while (root != null || (!stack.isEmpty())) {
if (root != null) {
stack.push(root);
root = root.left;
} else {
root = stack.pop();
if (prev != null) {
min = Math.min(min, root.val - prev.val);
}
prev = root;
root = root.right;
}
}

return min;
}
}

源码分析

递归的解法中需要特别注意 min 和 prev 的设定,作为参数传入均不太妥当。由于二叉搜索树的特性,求得最小差值时无需判断绝对值。

复杂度分析

递归实现用了隐式栈,迭代实现用了显式栈,最坏情况下栈的大小均为节点总数,平均情况下为树的高度。故平均情况下空间复杂度为 \[O(\log n)\], 每个节点被访问一次,时间复杂度为 \[O(n)\].

Reference




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