783. Minimum Distance Between BST Nodes
2018-03-21 10:42
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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.
Example :
Note:
The size of the BST will be between 2 and 100.
The BST is always valid, each node’s value is an integer, and each node’s value is different.
思路:
BST + 中序,再求minimum difference
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:
The size of the BST will be between 2 and 100.
The BST is always valid, each node’s value is an integer, and each node’s value is different.
思路:
BST + 中序,再求minimum difference
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int minDiffInBST(TreeNode root) {
vis = new ArrayList<Integer>();
dfs(root);
int min = 0x3f3f3f3f;
for (int i = 1; i < vis.size(); ++i) {
int diff = vis.get(i) - vis.get(i - 1);
min = Math.min(min, diff);
}
return min;
}
List<Integer> vis;
public void dfs(TreeNode root) {
if (root == null) return;
dfs(root.left);
vis.add(root.val);
dfs(root.right);
}
}
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