783. 二叉搜索树节点最小距离

题目描述

给你一个二叉搜索树的根节点 root ，返回 树中任意两不同节点值之间的最小差值 。

差值是一个正数，其数值等于两值之差的绝对值。

 

示例 1：

输入：root = [4,2,6,1,3]
输出：1

示例 2：

输入：root = [1,0,48,null,null,12,49]
输出：1

 

提示：

• 树中节点的数目范围是 [2, 100]
• 0 <= Node.val <= 105

 

注意：本题与 530：https://leetcode.cn/problems/minimum-absolute-difference-in-bst/ 相同

解法

方法一：中序遍历

题目需要我们求任意两个节点值之间的最小差值，而二叉搜索树的中序遍历是一个递增序列，因此我们只需要求中序遍历中相邻两个节点值之间的最小差值即可。

我们可以使用递归的方法来实现中序遍历，过程中用一个变量 \(\textit{pre}\) 来保存前一个节点的值，这样我们就可以在遍历的过程中求出相邻两个节点值之间的最小差值。

时间复杂度 \(O(n)\)，空间复杂度 \(O(n)\)。其中 \(n\) 为二叉搜索树的节点个数。

Python3JavaC++GoTypeScriptRustJavaScript

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def minDiffInBST(self, root: Optional[TreeNode]) -> int:
        def dfs(root: Optional[TreeNode]):
            if root is None:
                return
            dfs(root.left)
            nonlocal pre, ans
            ans = min(ans, root.val - pre)
            pre = root.val
            dfs(root.right)

        pre = -inf
        ans = inf
        dfs(root)
        return ans

/**
 * 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 {
    private final int inf = 1 << 30;
    private int ans = inf;
    private int pre = -inf;

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

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        dfs(root.left);
        ans = Math.min(ans, root.val - pre);
        pre = root.val;
        dfs(root.right);
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int minDiffInBST(TreeNode* root) {
        const int inf = 1 << 30;
        int ans = inf, pre = -inf;
        auto dfs = [&](this auto&& dfs, TreeNode* root) -> void {
            if (!root) {
                return;
            }
            dfs(root->left);
            ans = min(ans, root->val - pre);
            pre = root->val;
            dfs(root->right);
        };
        dfs(root);
        return ans;
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func minDiffInBST(root *TreeNode) int {
    const inf int = 1 << 30
    ans, pre := inf, -inf
    var dfs func(*TreeNode)
    dfs = func(root *TreeNode) {
        if root == nil {
            return
        }
        dfs(root.Left)
        ans = min(ans, root.Val-pre)
        pre = root.Val
        dfs(root.Right)
    }
    dfs(root)
    return ans
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function minDiffInBST(root: TreeNode | null): number {
    let [ans, pre] = [Infinity, -Infinity];
    const dfs = (root: TreeNode | null) => {
        if (!root) {
            return;
        }
        dfs(root.left);
        ans = Math.min(ans, root.val - pre);
        pre = root.val;
        dfs(root.right);
    };
    dfs(root);
    return ans;
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    pub fn min_diff_in_bst(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        const inf: i32 = 1 << 30;
        let mut ans = inf;
        let mut pre = -inf;

        fn dfs(node: Option<Rc<RefCell<TreeNode>>>, ans: &mut i32, pre: &mut i32) {
            if let Some(n) = node {
                let n = n.borrow();
                dfs(n.left.clone(), ans, pre);
                *ans = (*ans).min(n.val - *pre);
                *pre = n.val;
                dfs(n.right.clone(), ans, pre);
            }
        }

        dfs(root, &mut ans, &mut pre);
        ans
    }
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var minDiffInBST = function (root) {
    let [ans, pre] = [Infinity, -Infinity];
    const dfs = root => {
        if (!root) {
            return;
        }
        dfs(root.left);
        ans = Math.min(ans, root.val - pre);
        pre = root.val;
        dfs(root.right);
    };
    dfs(root);
    return ans;
};

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