Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree [1,null,2,3],
1
\
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
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解答
原来那个returnSize是拿来返回产生的中序遍历的数组大小用的……数组下标从0开始计算……
/**
* 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) {
], top = -, i = ;
);
!= top||NULL != root){
while(NULL != root){
stack[++top] = root;
root = root->left;
}
root = stack[top--];
return_array[i++] = root->val;
root = root->right;
}
*returnSize = i;
return return_array;
}