每天一道LeetCode-----寻找二叉搜索树中第k小的元素

Kth Smallest Element in a BST

原题链接Kth Smallest Element in a BST

给顶一个二叉搜索树的根节点，找到这棵数第k小的值

两种方法

• 递归法的中序遍历
• 迭代法的中序遍历

递归法，常规的中序遍历

代码如下

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int kthSmallest(TreeNode* root, int k) {        int res = 0;        inOrder(root, k, res);        return res;    }private:    void inOrder(TreeNode* root, int& k, int& res)    {        if(!root || !k) return;        inOrder(root->left, k, res);        //找到k个数即可        if(--k == 0)            res = root->val;        inOrder(root->right, k, res);    }};

迭代法，用栈实现

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int kthSmallest(TreeNode* root, int k) {        stack<TreeNode*> s;        pushAllLeft(root, s);        TreeNode* top = nullptr;        while(k--)        {            top = s.top();            s.pop();            //每弹出一个节点就将该节点的右子树的左边一列入栈            pushAllLeft(top->right, s);        }        return top->val;    }private:    void pushAllLeft(TreeNode*root, stack<TreeNode*>& s)    {        while(root)        {            s.push(root);            root = root->left;        }    }};

本题主要考察中序遍历，除了递归法最好可以熟悉迭代法的求解