【剑指offer】【树】54.二叉搜索树的第k大节点

二叉搜索树的第k大节点

递归法

中序遍历的二叉搜索树序列为单调递增的序列，将中序遍历的结果放到vector中，第k大的数为v.size()-k位置的数

/**
 * 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> v;
    void dfs(TreeNode * root)
    {
        if(root->left) dfs(root->left);
        v.push_back(root->val);
        if(root->right) dfs(root->right);
    }
    int kthLargest(TreeNode* root, int k) {
        dfs(root);
        return v[v.size() - k];
    }
};

迭代法

/**
 * 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 kthLargest(TreeNode* root, int k) {
        stack<TreeNode *> st;
        int i = 0;
        while(!st.empty() || root)
        {
            while(root)
            {
                st.push(root);
                root = root -> right;
            }
            root = st.top();
            st.pop();
            i++ ;
            if(i == k) return root->val;
            root = root->left;
        }
        return 0;
    }
};

原文地址：https://www.cnblogs.com/Trevo/p/12682819.html