Binary Tree Preorder Traversal leetcode144

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?

先贴出递归的解法，但提交时会出现runtime error

/**
* 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!=NULL)result.push_back(root->val);
if(root->left!=NULL)preorderTraversal(root->left);
if(root->right!=NULL)preorderTraversal(root->right);
return result;
}
};

不用递归用迭代的话，就必须要借助栈来保存访问过的节点，以便访问其右孩子。

/**
* 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 *> s;
TreeNode* p;
while(p!=NULL||!s.empty()){
while(p!=NULL){
result.push_back(p->val);
s.push(p);//将访问过的节点压栈，以便后面访问其右孩子
p=p->left;//访问其左孩子
}
if(!s.empty()){//此时p节点为NULL，若栈不为空的话，则把p指向先前访问过的节点，访问该节点的右孩子
p=s.top();
s.pop();
p=p->right;
}
}
return result;
}
};

Binary Tree Preorder Traversal

Submission Details

67 / 67 test cases passed.

Status: Accepted

Runtime: 3 ms

Submitted: 45 minutes ago