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144. Binary Tree Preorder Traversal
Problem Statement
Given a binary tree, return the preorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3},
1 2 / 3
return [1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?
solution one:
traverse:
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */10 class Solution {11 public:12 void preorderTraverse(TreeNode *node, vector<int>& preorder){13 if(node == NULL){14 return;15 }16 17 preorder.push_back(node->val);18 preorderTraverse(node->left, preorder);19 preorderTraverse(node->right, preorder);20 return;21 }22 23 vector<int> preorderTraversal(TreeNode* root) {24 vector<int> preorder;25 preorderTraverse(root, preorder);26 return preorder; 27 }28 };
Divide and conquer:
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */10 class Solution {11 public:12 // this version is divide && conquer13 vector<int> preorderTraversal(TreeNode* root) {14 vector<int> preorder;15 if(root == NULL){16 return preorder;17 }18 // divide19 vector<int> left = preorderTraversal(root->left);20 vector<int> right = preorderTraversal(root->right);21 // conquer22 preorder.push_back(root->val);23 preorder.insert(preorder.end(), left.begin(), left.end());24 preorder.insert(preorder.end(), right.begin(), right.end());25 return preorder;26 }27 };
144. Binary Tree Preorder Traversal
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