LintCode 二叉树的前序遍历
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给出一棵二叉树,返回其节点值的前序遍历。
样例
给出一棵二叉树 {1,#,2,3}
,
1
2
/
3
返回 [1,2,3]
.
你能使用非递归实现么?
分析:使用非递归实现(栈)
* 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 TreeNode *curr=root; stack<TreeNode *> mystack; vector<int> res; while(!mystack.empty()||curr!=NULL) { while(curr!=NULL) { res.push_back(curr->val); mystack.push(curr); curr=curr->left; } if(!mystack.empty()) { curr=mystack.top(); mystack.pop(); curr=curr->right; } } return res; } };
还可以用数组指针。
/** * 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 TreeNode *curr=root; TreeNode *mystack[1000]; int top=0; vector<int> res; while(top!=0||curr!=NULL) { while(curr!=NULL) { res.push_back(curr->val); mystack[top++]=curr; curr=curr->left; } if(top>0) { top--; curr=mystack[top]; curr=curr->right; } } return res; } };
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