剑指offer 54、二叉搜索树的第k大节点

方法1：中序反向遍历，找到后还会继续遍历

class Solution {
public:
    int kthLargest(TreeNode* root, int k) {
        int result = 0;
        dfs(root, result, k);
        return result;
    }

private:
    void dfs(TreeNode *root, int &result, int &k) {
        if (!root) return;
        dfs(root->right, result, k);
        if (!--k) result = root->val;
        dfs(root->left, result, k);
    }
};

方法2：遍历到后就停止

class Solution {
public:
    int kthLargest(TreeNode* root, int &k) {
        if (!root) return 0;
        int result = kthLargest(root->right, k);
        if (!--k) return root->val;
        return k < 0 ? result : kthLargest(root->left, k);
    }
};

/**
 * 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 {
private:
    int res;
    int k;
public:
    int kthLargest(TreeNode* root, int k) {
        this->k = k;
        dfs(root);
        return res;
    }

    void dfs(TreeNode* curr)
    {
        if (curr != nullptr)
        {
            dfs(curr->right);
            // 正好是第k个，则更新res
            if (--k == 0)
            {
                res = curr->val;
                return;
            }
            dfs(curr->left);
        }
    }
};

方法3：非递归方式

class Solution {
public:
    int kthLargest(TreeNode* root, int k) {
        vector<TreeNode*> worker;
        while (root || worker.size()) {
            while (root) {
                worker.push_back(root); // 根入栈
                root = root->right; // 访问右子树，向下探
            }
            root = worker.back(), worker.pop_back(); // 出栈
            if (!--k) return root->val;
            root = root->left; // 访问左子树
        }
        return 0;
    }
};