After writing such a question on Leetcode, I feel that my thinking is quite good, share it:
Idea: First, add each non-null node val to a set through depth-first search, and then traverse the set to find the minimum value of the difference.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private int min = Integer.MAX_VALUE;
public int minDiffInBST(TreeNode root) {
List<Integer> valueList = new ArrayList<Integer>();
dfs(root,valueList);
for(int i = 0;i < valueList.size();i++){
for(int j = i + 1;j < valueList.size();j++){
min = Math.min(min,Math.abs(valueList.get(i)-valueList.get(j)));
}
}
return min;
}
public void dfs(TreeNode node,List<Integer> list){
if(node != null){
int a = node.val;
list.add(a);
dfs(node.left,list);
dfs(node.right,list);
}
}
}But it is obvious: the double for loop greatly increases the time complexity!
Then, we should think about how to optimize:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private int min = Integer.MAX_VALUE;
public int minDiffInBST(TreeNode root) {
List<Integer> valueList = new ArrayList<Integer>();
dfs(root,valueList);
/*for(int i = 0;i < valueList.size();i++){
for(int j = i + 1;j < valueList.size();j++){
min = Math.min(min,Math.abs(valueList.get(i)-valueList.get(j)));
}
}*/
Collections.sort(valueList);
for(int i = 0; i < valueList.size() - 1;i++){
min = Math.min(min,valueList.get(i+1)-valueList.get(i));
}
return min;
}
public void dfs(TreeNode node,List<Integer> list){
if(node != null){
int a = node.val;
list.add(a);
dfs(node.left,list);
dfs(node.right,list);
}
}
}Let's take a look at the effect:
