lintcode 11. 二叉查找樹中搜索區間

 

中序遍歷一遍就好啦

/**
 * 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: param root: The root of the binary search tree
     * @param k1: An integer
     * @param k2: An integer
     * @return: 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>ans;
        if(root==NULL) return ans;
        TreeNode* p = root;
        stack<TreeNode*>s;
        while(!s.empty()||p) {
            if(p) {
                s.push(p);
                p = p->left;
            } else {
                p = s.top();
                s.pop();
                if (p->val >= k1 && p->val <= k2) {
                    ans.push_back(p->val);
                }
                p = p->right;
            }
        }
        return ans;
    }
};

其實比較好的 遞歸的時候 不滿足大小範圍的子樹就不遍歷了。。。

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