二叉搜索树的最小绝对差

本文参考代码随想录

给你一棵所有节点为非负值的二叉搜索树，请你计算树中任意两节点的差的绝对值的最小值。

递归法1

递归法中序遍历二叉搜索树，得到有序数组，然后遍历一遍数组

class Solution {
private:
    vector<int> vec;
    void traversal(TreeNode* root) {
        if(root == nullptr) return;
        traversal(root->left);
        vec.push_back(root->val);
        traversal(root->right);
    }
public:
    int getMinimumDifference(TreeNode* root) {
        vec.clear();
        traversal(root);
        if(vec.size() < 2) return 0;
        int result = INT_MAX;
        for(int i = 1;i < vec.size(); i++) {
            result = min(result, vec[i] - vec[i - 1]);
        }
        return result;
    }
};

递归法2

在二叉树遍历过程中直接计算，记录当前节点的前一个节点pre

class Solution {
private:
    int result = INT_MAX;
    TreeNode* pre = nullptr;
    void traversal(TreeNode* cur) {
        if(cur == nullptr) return;
        traversal(cur->left);
        if(pre != nullptr) {
            result = min(result, cur->val - pre->val);
        }
        pre = cur;
        traversal(cur->right);
    }
public:
    int getMinimumDifference(TreeNode* root) {
        traversal(root);
        return result;
    }
};

迭代法

class Solution {
public:
    int getMinimumDifference(TreeNode* root) {
        stack<TreeNode*> st;
        TreeNode* cur = root;
        TreeNode* pre = nullptr;
        int result = INT_MAX;
        while(cur != nullptr || !st.empty()) {
            if(cur != nullptr) {
                st.push(cur);
                cur = cur->left;
            }else{
                cur = st.top();
                st.pop();
                if(pre != nullptr) {
                    result = min(result, cur->val - pre->val);
                }
                pre = cur;
                cur = cur->right;
            }
        }
        return result;
    }
    
};