二叉查找树中搜索区间

给定两个值 k1 和 k2（k1 < k2）和一个二叉查找树的根节点。找到树中所有值在 k1 到 k2 范围内的节点。即打印所有x (k1 <= x <= k2) 其中 x 是二叉查找树的中的节点值。返回所有升序的节点值。

如果有 k1 =

10

和 k2 =

22

,
你的程序应该返回

[12, 20, 22]

.

20
/  \
8   22
/ \
4   12

/**
* Definition of TreeNode:
* class TreeNode {
* public:
*     int val;
*     TreeNode *left, *right;
*     TreeNode(int val) {
*         this->val = val;
*         this->left = this->right = NULL;
*     }
* }
*/
class Solution {
public:
/**
* @param root: The root of the binary search tree.
* @param k1 and k2: range k1 to k2.
* @return: Return all keys that k1<=key<=k2 in ascending order.
*/
vector<int> searchRange(TreeNode* root, int k1, int k2) {
// write your code here
vector<int> res;
helper(root,k1,k2,res);
return res;
}
void helper(TreeNode* root,int k1,int k2,vector<int>&res){
if(root==NULL) return;
if(root->val<=k2&&root->val>=k1){
helper(root->left,k1,k2,res);
res.push_back(root->val);
helper(root->right,k1,k2,res);
}
else if(root->val>k2) helper(root->left,k1,k2,res);
else helper(root->right,k1,k2,res);
}
};