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501. 二叉搜索树中的众数

题目描述

给你一个含重复值的二叉搜索树（BST）的根节点 root ，找出并返回 BST 中的所有 众数（即，出现频率最高的元素）。

如果树中有不止一个众数，可以按 任意顺序 返回。

假定 BST 满足如下定义：

  • 结点左子树中所含节点的值 小于等于 当前节点的值
  • 结点右子树中所含节点的值 大于等于 当前节点的值
  • 左子树和右子树都是二叉搜索树

 

示例 1：

输入：root = [1,null,2,2]
输出：[2]

示例 2：

输入：root = [0]
输出：[0]

 

提示：

  • 树中节点的数目在范围 [1, 104] 内
  • -105 <= Node.val <= 105

 

进阶：你可以不使用额外的空间吗？（假设由递归产生的隐式调用栈的开销不被计算在内）

解法

方法一

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# 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 findMode(self, root: TreeNode) -> List[int]:
        def dfs(root):
            if root is None:
                return
            nonlocal mx, prev, ans, cnt
            dfs(root.left)
            cnt = cnt + 1 if prev == root.val else 1
            if cnt > mx:
                ans = [root.val]
                mx = cnt
            elif cnt == mx:
                ans.append(root.val)
            prev = root.val
            dfs(root.right)

        prev = None
        mx = cnt = 0
        ans = []
        dfs(root)
        return ans

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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 {
    private int mx;
    private int cnt;
    private TreeNode prev;
    private List<Integer> res;

    public int[] findMode(TreeNode root) {
        res = new ArrayList<>();
        dfs(root);
        int[] ans = new int[res.size()];
        for (int i = 0; i < res.size(); ++i) {
            ans[i] = res.get(i);
        }
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        dfs(root.left);
        cnt = prev != null && prev.val == root.val ? cnt + 1 : 1;
        if (cnt > mx) {
            res = new ArrayList<>(Arrays.asList(root.val));
            mx = cnt;
        } else if (cnt == mx) {
            res.add(root.val);
        }
        prev = root;
        dfs(root.right);
    }
}

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/**
 * 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:
    TreeNode* prev;
    int mx, cnt;
    vector<int> ans;

    vector<int> findMode(TreeNode* root) {
        dfs(root);
        return ans;
    }

    void dfs(TreeNode* root) {
        if (!root) return;
        dfs(root->left);
        cnt = prev != nullptr && prev->val == root->val ? cnt + 1 : 1;
        if (cnt > mx) {
            ans.clear();
            ans.push_back(root->val);
            mx = cnt;
        } else if (cnt == mx)
            ans.push_back(root->val);
        prev = root;
        dfs(root->right);
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findMode(root *TreeNode) []int {
    mx, cnt := 0, 0
    var prev *TreeNode
    var ans []int
    var dfs func(root *TreeNode)
    dfs = func(root *TreeNode) {
        if root == nil {
            return
        }
        dfs(root.Left)
        if prev != nil && prev.Val == root.Val {
            cnt++
        } else {
            cnt = 1
        }
        if cnt > mx {
            ans = []int{root.Val}
            mx = cnt
        } else if cnt == mx {
            ans = append(ans, root.Val)
        }
        prev = root
        dfs(root.Right)
    }
    dfs(root)
    return ans
}

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public class Solution {
    private int mx;
    private int cnt;
    private TreeNode prev;
    private List<int> res;

    public int[] FindMode(TreeNode root) {
        res = new List<int>();
        Dfs(root);
        int[] ans = new int[res.Count];
        for (int i = 0; i < res.Count; ++i) {
            ans[i] = res[i];
        }
        return ans;
    }

    private void Dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        Dfs(root.left);
        cnt = prev != null && prev.val == root.val ? cnt + 1 : 1;
        if (cnt > mx) {
            res = new List<int>(new int[] { root.val });
            mx = cnt;
        } else if (cnt == mx) {
            res.Add(root.val);
        }
        prev = root;
        Dfs(root.right);
    }
}

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