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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?
递归方式很简单：
1 vector<int> preorderTraversal(TreeNode *root) 2 { 3 vector<int> vtree; 4 preorder(vtree, root); 5 return vtree; 6 } 7 8 void preorder(vector<int> &vtree, TreeNode *root) 9 { 10 if (root == NULL) 11 return; 12 vtree.push_back(root->val); 13 preorder(vtree, root->left); 14 preorder(vtree, root->right); 15 }
对于非递归方式，需要借助堆栈来临时存储tree node
1) if current node P is not empty, visit it and put P->right in stack, then P = P->left
2) else, get the top node in stack
1 vector<int> preorderTraversal(TreeNode *root) 2 { 3 vector<int> vtree; 4 stack<TreeNode *> stree; 5 TreeNode *visit = root; 6 7 while (!stree.empty() || (visit != NULL)) 8 { 9 if (visit != NULL) 10 { 11 vtree.push_back(visit->val); // if visit is not empty, visit it 12 if (visit->right != NULL) // if visit->right exit, push it in stack 13 stree.push(visit->right); 14 visit = visit->left; 15 } 16 else // visit is null 17 { 18 visit = stree.top(); 19 stree.pop(); 20 } 21 } 22 23 return vtree; 24 }
leetcode 144. Binary Tree Preorder Traversal
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原文地址：http://www.cnblogs.com/ym65536/p/4231519.html
