2015年9月9日星期三

Lintcode: Binary Tree Preorder Traversal Show result


Recursive solution:
Time: O(n) -- n number of nodes
Space: O(L) -- tree height

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/**
 * 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 binary tree.
     * @return: Preorder in vector which contains node values.
     */
    vector<int> preorderTraversal(TreeNode *root) {
        // write your code here
        vector<int> result;
        preorder(root, result);
        return result;
    }
    void preorder(TreeNode* node, vector<int>& v){
        if(!node) return;
        v.push_back(node->val);
        if(node->left) preorder(node->left, v);
        if(node->right) preorder(node->right, v);
    }
};

Iterative solution:
Use stack to record nodes, remember push the right child first then left child ---- make sure pop in post order

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/**
 * 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 binary tree.
     * @return: Preorder in vector which contains node values.
     */
    vector<int> preorderTraversal(TreeNode *root) {
        vector<int> result;
        if(!root) return result; 
        stack<TreeNode*> s;
        s.push(root);
        while(!s.empty()){
            TreeNode *cur = s.top();
            s.pop();
            result.push_back(cur->val);
            if(cur->right) s.push(cur->right);
            if(cur->left) s.push(cur->left);
        }
        return result;
    }
};
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