leetcode[94]:Binary Tree Inorder Traversal

Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes’ values.

For example:

Given binary tree {1,#,2,3},

1
\
2
/
3

return [1,3,2].

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* inorderTraversal(struct TreeNode* root, int* returnSize) {
struct TreeNode stack[1000];
struct TreeNode *tmp,*tmp2;
int i=0;
int res[100000]={0},*result;
int k=0;
if(!root) {*returnSize=k;return NULL;}

tmp=root;

while(1)
{

if(tmp->left)
{
stack[i++] = *tmp;
stack[i-1].left=NULL;
tmp=tmp->left;
continue;
}
res[k++] = tmp->val;
if(tmp->right)
{
tmp=tmp->right;
continue;
}

if(i==0) break;
tmp = &stack[--i];
}
* returnSize =k;
result= (int*)malloc(sizeof(int*)*k);
for(i=0;i<k;i++)
{
result[i]=res[i];
}
return result;
}

有左子树，指向他，并将根节点左赋空。