标签：des blog io strong for ar div amp
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?
1st （ 8 tries）
/**
* Definition for binary tree
* 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)
{
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
vector<int> re;
/*
preorder(root,re);
return re;
*/
/*iteratively*/
stack<TreeNode *> st;
if(root == NULL)
return re;
re.push_back(root->val);
st.push(root);
TreeNode *iter = root->left;
/*this is important,while condition*/
while(!st.empty() || /*this is import*/iter != NULL)
{
while(iter != NULL)
{
re.push_back(iter->val);
st.push(iter);
iter = iter->left;
}
iter = st.top();
st.pop();
iter = iter->right;
}
return re;
}
/*
void preorder(TreeNode *root,vector<int> &re)
{
if(root == NULL)
return;
re.push_back(root->val);
preorder(root->left,re);
preorder(root->right,re);
}
*/
};
2nd （ 6 tries）
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> ans;
vector<int> preorderTraversal(TreeNode *root) {
//recursive
/*
recursive(root);
return ans;
*/
//iteratively
stack<TreeNode *> s;
if(root == NULL)
return ans;
TreeNode *cur = root;
//first check iterative pointor is not NULL or stack is not NULL,
//it imply that there is some node you can scan
while( cur != NULL || !s.empty() ) {
//iterative all the left node
while(cur != NULL) {
ans.push_back(cur->val);
s.push(cur);
cur = cur->left;
}
//if all left node scaned,goto the right child tree
cur = s.top();
s.pop();
cur = cur->right;
}
return ans;
}
/*
void recursive(TreeNode *root) {
//step 1
if(root == NULL)
return;
//step 2
//step 3
ans.push_back(root->val);
recursive(root->left);
recursive(root->right);
}
*/
};
/**
* Definition for binary tree
* 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> ans;
TreeNode *cur = NULL,*pre = NULL;
cur = root;
while( cur ) {
if( cur->left == NULL ) {
//in-order here!!!
//pre-order here!!!
ans.push_back(cur->val);
cur = cur->right;
}
else {
pre = cur->left;
while( pre->right != NULL && pre->right != cur ) {
pre = pre->right;
}
if( pre->right == NULL ) {
pre->right = cur;
//pre-order here!!!
ans.push_back(cur->val);
cur = cur->left;
}
else {
pre->right = NULL;
//in-order here!!!
cur = cur->right;
}
}
}
return ans;
}
};
【Leetcode】Binary Tree Preorder Traversal,布布扣,bubuko.com
【Leetcode】Binary Tree Preorder Traversal
标签：des blog io strong for ar div amp
原文地址：http://www.cnblogs.com/weixliu/p/3924397.html
