标签：turn nod cto value 节点 values 前序遍历 traversal accordion
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;
}
};
标签：turn nod cto value 节点 values 前序遍历 traversal accordion
原文地址：http://www.cnblogs.com/lelelelele/p/6115274.html
