783 Minimum Distance Between BST Nodes
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.
(1) Inorder Traversal
/**
* 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) {
int res = Integer.MAX_VALUE;
TreeNode curNode = root, preNode = null;
Stack<TreeNode> stack = new Stack<>();
while (curNode != null || !stack.isEmpty()) {
if (curNode != null) {
stack.push(curNode);
curNode = curNode.left;
}
else {
curNode = stack.pop();
if (preNode != null) {
res = Math.min(res, Math.abs(preNode.val - curNode.val));
}
preNode = curNode;
curNode = curNode.right;
}
}
return res;
}
}
时间复杂度O(n), n为树的node的个数，空间复杂度O(h), h为树的高度
Last modified 3yr ago