LC.P783[二叉搜索树节点最小距离]

方法一：遍历+排序

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/**
 * 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) {
        List<Integer> list = new ArrayList<>();
        dfs(root, list);
        list.sort((a, b) -> a - b);
        int ans = Integer.MAX_VALUE;
        for (int i = 1; i < list.size(); ++i) {
            ans = Math.min(ans, Math.abs(list.get(i - 1) - list.get(i)));
        }
        return ans;
    }

    private void dfs(TreeNode root, List<Integer> list) {
        if (root == null) return;
        list.add(root.val);
        dfs(root.left, list);
        dfs(root.right, list);
    }
}
  • 时间复杂度：$O(nlogn)$
  • 空间复杂度：$O(n)$

方法二：中序遍历

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/**
 * 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 {
    TreeNode pre = null;
    int ans = Integer.MAX_VALUE;

    public int minDiffInBST(TreeNode root) {
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null) return;
        dfs(root.left);
        if (pre != null) {
            ans = Math.min(ans, Math.abs(root.val - pre.val));
        }
        pre = root;
        dfs(root.right);
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$