LeetCode.783 二叉搜索树节点最小距离

原题

https://leetcode-cn.com/problems/minimum-distance-between-bst-nodes/

思路

利用中序遍历递增的性质

• （1）中序遍历得到一个递增的集合，遍历集合得到

  min

• （2）遍历过程中记录

  pre

  ，直接得到

  min

题解

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int minDiffInBST(TreeNode root) {
//        中序遍历计数
        List<Integer> list = new ArrayList<>();
        midOrder(root, list);

        int len = list.size();
        int res = Integer.MAX_VALUE;
        for (int i = 1; i < len; i++) {
            res = Math.min(res, list.get(i) - list.get(i-1));
        }

        return res;
    }

    private static void midOrder(TreeNode node, List<Integer> list){
        if (node.left != null) {
            midOrder(node.left, list);
        }
        list.add(node.val);
        if (node.right != null) {
            midOrder(node.right, list);
        }
    }

}
           

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {

    int pre;
    int min;

    public int minDiffInBST(TreeNode root) {
        min = Integer.MAX_VALUE;
        pre = -1;
        midOrder2(root);
        return min;
    }

    private void midOrder2(TreeNode node){
        if (node.left != null) {
            midOrder2(node.left);
        }
        if (pre == -1) {
            pre = node.val;
        } else {
            min = Math.min(min, node.val - pre);
            pre = node.val;
        }

        if (node.right != null) {
            midOrder2(node.right);
        }
    }

}