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~~~

  1 /* inorder traversal binary tree */
  2 #include <stdio.h>
  3 #include <stdlib.h>
  4
  5
  6 struct TreeNode {
  7     int val;
  8     struct TreeNode *left;
  9     struct TreeNode *right;
 10 };
 11
 12 int* inorderTraversal(struct TreeNode* root, int* returnSize);
 13
 14 int main() {
 15
 16     struct TreeNode n1, n2, n3;
 17     n1.val = 1;
 18     n1.left = NULL;
 19     n1.right = &n2;
 20     /*  */
 21     n2.val = 2;
 22     n2.left = &n3;
 23     n2.right = NULL;
 24     /*  */
 25     n3.val = 3;
 26     n3.left = NULL;
 27     n3.right = NULL;
 28     int returnSize = 0;
 29     int *a = inorderTraversal(&n1, &returnSize);
 30
 31     int i=0;
 32     for(i=0; i<returnSize; i++)
 33         printf("%d ", a[i]);
 34
 35     printf("\n");
 36
 37 }
 38
 39
 40 /**
 41  * Definition for a binary tree node.
 42  * struct TreeNode {
 43  *     int val;
 44  *     struct TreeNode *left;
 45  *     struct TreeNode *right;
 46  * };
 47  */
 48 /**
 49  * Return an array of size *returnSize.
 50  * Note: The returned array must be malloced, assume caller calls free().
 51  */
 52 int* inorderTraversal(struct TreeNode* root, int* returnSize) {
 53
 54     struct TreeNode  **stack = (struct TreeNode **) malloc (sizeof(struct TreeNode *) * 1000); /* store node not access at present */
 55     int *result = (int *) malloc(sizeof(int) * 1000);
 56     int count = 0;
 57     stack[0] = root;           /* first node */
 58     struct TreeNode* curr;
 59     int top = 0;                /* element number in stack */
 60
 61     while(top != -1 ) {
 62         curr = stack[top];    /* get stack top element */
 63
 64         if(curr == NULL) {     /* if current element is null */
 65             while(top != -1 && curr == NULL)
 66                 curr = stack[--top];
 67             if(top == -1 || curr == NULL)
 68                 break;
 69             else {
 70                 result[count++] = curr->val;
 71                 stack[top] = curr->right;
 72                 continue;
 73             }
 74         }
 75         if(curr->left != NULL)
 76             stack[++top] = curr->left;
 77
 78         if(curr->left == NULL) {  /* if left subtree is NULL, then we need to access middle node */
 79             result[count++] = curr->val;
 80             stack[top] = curr->right;
 81         }
 82     }
 83     *returnSize = count;
 84     return result;
 85 }