LeetCode：二叉树的非递归中序遍历

第一次动手写二叉树的，有点小激动，64行的if花了点时间，上传leetcode一次点亮~~~

 /* inorder traversal binary tree */
 #include <stdio.h>
 #include <stdlib.h>

 struct TreeNode {
     int val;
     struct TreeNode *left;
     struct TreeNode *right;
 };

 int* inorderTraversal(struct TreeNode* root, int* returnSize);

 int main() {

     struct TreeNode n1, n2, n3;
     n1.val = 1;
     n1.left = NULL;
     n1.right = &n2;
     /*  */
     n2.val = 2;
     n2.left = &n3;
     n2.right = NULL;
     /*  */
     n3.val = 3;
     n3.left = NULL;
     n3.right = NULL;
     int returnSize = 0;
     int *a = inorderTraversal(&n1, &returnSize);

     int i=0;
     for(i=0; i<returnSize; i++)
         printf("%d ", a[i]);

     printf("\n");

 }

 /**
  * 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 = (struct TreeNode **) malloc (sizeof(struct TreeNode *) * 1000); /* store node not access at present */
     int *result = (int *) malloc(sizeof(int) * 1000);
     int count = 0;
     stack[0] = root;           /* first node */
     struct TreeNode* curr;
     int top = 0;                /* element number in stack */

     while(top != -1 ) {
         curr = stack[top];    /* get stack top element */

         if(curr == NULL) {     /* if current element is null */
             while(top != -1 && curr == NULL)
                 curr = stack[--top];
             if(top == -1 || curr == NULL)
                 break;
             else {
                 result[count++] = curr->val;
                 stack[top] = curr->right;
                 continue;
             }
         }
         if(curr->left != NULL)
             stack[++top] = curr->left;

         if(curr->left == NULL) {  /* if left subtree is NULL, then we need to access middle node */
             result[count++] = curr->val;
             stack[top] = curr->right;
         }
     }
     *returnSize = count;
     return result;
 }