Binary Tree Preorder Traversal
Binary Tree Preorder Traversal
题目
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?
思路
先序遍历，利用栈和递归
解题
c++ 1版1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32/**
* 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:
vector<int> preorderTraversal(TreeNode* root) {
vector<int> result;
if(!root)
return result;
stack<TreeNode*> myStack;
TreeNode *cur=root;
while(cur || !myStack.empty()){
while(cur){
myStack.push(cur);
result.push_back(cur->val);
cur=cur->left;
}
if(!myStack.empty()){
cur=myStack.top()->right;
myStack.pop();
}
}
return result;
}
};
c++ 2版1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27/**
* 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:
vector<int> preorderTraversal(TreeNode* root) {
vector<int> result;
stack<TreeNode*> ss;
ss.push(root);
while(!ss.empty()){
TreeNode* cur=ss.top();
ss.pop();
if(cur!=NULL){
result.push_back(cur->val);
ss.push(cur->right);
ss.push(cur->left);
}
}
return result;
}
};
python版1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24# Definition for a binary tree node
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# @param root, a tree node
# @return a list of integers
def preorderTraversal(self, root):
stack=[]
T=root
result=[]
while T or (len(stack)!=0):
while T:
result.append(T.val)
stack.append(T)
T=T.left
if(len(stack)!=0):
tmp=stack[-1]
stack.pop()
T=tmp.right
return result
