LeetCode鏈接:https://leetcode-cn.com/problems/binary-tree-preorder-traversal/
給定一個二叉樹,返回它的 前序 遍歷。
示例:
輸入: [1,null,2,3]
1
\
2
/
3
輸出: [1,2,3]
一,非遞歸解決方案:
根據前序遍歷訪問的順序,優先訪問根結點,然後再分別訪問左孩子和右孩子。即對於任一結點,其可看做是根結點,因此可以直接訪問,訪問完之後,若其左孩子不爲空,按相同規則訪問它的左子樹;當訪問其左子樹時,再訪問它的右子樹。因此其處理過程如下:
對於任一結點cur:
1)訪問結點cur,並將結點cur入棧;
2)判斷結點cur的左孩子是否爲空,若爲空,則取棧頂結點並進行出棧操作,並將棧頂結點的右孩子置爲當前的結點cur,循環至1);若不爲空,則將cur的左孩子置爲當前的結點cur;
3)直到cur爲NULL並且棧爲空,則遍歷結束。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
/*void preorder(TreeNode* root,vector<int>& v)
{
if(root == 0)
return;
v.push_back(root->val);
preorder(root->left,v);
preorder(root->right,v);
}*/
vector<int> preorderTraversal(TreeNode* root)
{
vector<int> v;
//preorder(root,v);
TreeNode* cur = root;
stack<TreeNode*> st;
while(cur || !st.empty())
{
while(cur)
{
st.push(cur);
v.push_back(cur->val);
cur = cur->left;
}
TreeNode* top = st.top();
st.pop();
cur = top->right;
}
return v;
}
};
二,進階: 遞歸算法很簡單,你可以通過迭代算法完成嗎?
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int *p;
int size;
void _pre(struct TreeNode *root){
if(root == NULL){
return;
}
p[size] = root->val;
size++;
_pre(root->left);
_pre(root->right);
}
int* preorderTraversal(struct TreeNode* root, int* returnSize) {
p = (int *)malloc(sizeof(int)*100000);
size = 0;
_pre(root);
*returnSize = size;
return p;
}
三,Python方案:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def preorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
if root == None:
return[]
return
[root.val]+self.preorderTraversal(root.left)+self.preorderTraversal(root.right)