Sword finger offer 54 The k-th node of binary search tree

Original link ： The finger of the sword Offer 54. The second of binary search tree k Big node

solution:

        The middle order traversal of binary search tree is an ascending array , Go back to k A big number is enough

/**
 * 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:
    vector<int> inorder;
    int kthLargest(TreeNode* root, int k) {
        dfs(root);
        return inorder[inorder.size() - k];
    }

    void dfs(TreeNode *root) {
        if(root == NULL) return;
        dfs(root->left);
        inorder.push_back(root->val);
        dfs(root->right);
    }
};

Reverse middle order traversal , In calculating the number of traversal nodes , Reduce memory consumption

/**
 * 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 ans = 0,cnt = 0;
    int kthLargest(TreeNode* root, int k) {
        dfs(root, k);
        return ans;
    }

    void dfs(TreeNode *root, int k) {
        if(root == NULL) return;

        dfs(root->right, k);
        if(++cnt == k) {
            ans = root->val;
            return;
        }
        dfs(root->left, k);
    }
};