leetcode[144]:Binary Tree Preorder Traversal

Binary Tree Preorder Traversal

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?

/**
* 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* preorderTraversal(struct TreeNode* root, int* returnSize) {
struct TreeNode stack[1000];
struct TreeNode *tmp;
int i=0;
int res[100000]={0},*result;
int k=0;
if(!root) {k+=0;*returnSize=k;
return NULL;}
tmp=root;

while(i!=-1)
{
res[k++] = tmp->val;
if(tmp->left && tmp->right)
{
stack[i++] = *tmp;
tmp=tmp->left;
continue;
}
if(tmp->left && !tmp->right)
{
tmp=tmp->left;
continue;
}
if(!tmp->left && tmp->right)
{
tmp=tmp->right;
continue;
}
tmp=&stack[--i];
tmp=tmp->right;
}
* returnSize =k;
result= (int*)malloc(sizeof(int*)*k);
for(i=0;i<k;i++)
{
result[i]=res[i];
}
return result;
}

左右子树都有需要保存节点，否则直接指向唯一节点。