題目描述:
給定一個二叉樹,傳回它的中序 周遊。
示例:
輸入: [1,null,2,3]
1
\
2
/
3
輸出: [1,3,2]
進階: 遞歸算法很簡單,你可以通過疊代算法完成嗎?
AC C++ Solution:
1.遞歸法:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
/* 遞歸法 */
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> nodes;
inorder(root,nodes);
return nodes;
}
private:
void inorder(TreeNode* root, vector<int>& nodes) {
if(!root)
return;
inorder(root->left,nodes); //當左子節點為空時,通路目前節點值
nodes.push_back(root->val);
inorder(root->right,nodes);
}
};
2.堆棧法:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
/* 堆棧法*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> vector;
stack<TreeNode *> stack;
TreeNode *pCurrent = root;
while(!stack.empty() || pCurrent)
{
if(pCurrent) {
stack.push(pCurrent);
pCurrent = pCurrent->left;
}
else {
pCurrent = stack.top();
stack.pop();
vector.push_back(pCurrent->val);
pCurrent = pCurrent->right;
}
}
return vector;
}
};
};