二叉树中第 K小的元素

给定一个二叉搜索树,编写一个函数 kthSmallest 来查找其中第 k 个最小的元素。

说明:
你可以假设 k 总是有效的,1 ≤ k ≤ 二叉搜索树元素个数。

思路: 二叉搜索树因其有序,故采用中序遍历,可以得到第K小的元素。

/**
 * 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 index = k;
        int kth = 0;
        TreeNode* find = findKth(root, index, kth);
        if(find) return find->val;
        return -1;
    }
    TreeNode* findKth(TreeNode* root, int index, int &k){
        if(root == NULL) {
            return NULL;
        }
        //左子树<根节点<右子树
        TreeNode* left = findKth(root->left,index, k);
        if(left) return left;
        k = k+1;
        if(k == index){  
           // cout << root->val << ",";
            return root;
        } 
        TreeNode* right = findKth(root->right, index, k);
        if(right) return right;
        else return NULL;        
        //return right? right:left;
    }
        
    
};
The Safest Way to Get what you Want is to Try and Deserve What you Want.
原文地址:https://www.cnblogs.com/Shinered/p/11498510.html