144. 二叉树的前序遍历

解1: 递归

void preorder(struct TreeNode* root, int *array, int *idx) {if (root != NULL) {array[(*idx)++] = root->val;preorder(root->left, array, idx);preorder(root->right, array, idx);}
}int* preorderTraversal(struct TreeNode* root, int* returnSize) {int *array = (int *)malloc(sizeof(int) * 100);memset(array, 0, sizeof(int) * 100);*returnSize = 0;preorder(root, array, returnSize);return array;
}

解2: 迭代

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     struct TreeNode *left;*     struct TreeNode *right;* };*/
/*** Note: The returned array must be malloced, assume caller calls free().*/struct stack{int top;struct TreeNode *stk[100];
};int* preorderTraversal(struct TreeNode* root, int* returnSize) {int *array = (int *)malloc(sizeof(int) * 100);struct stack *st = (struct stack *)malloc(sizeof(*st));struct TreeNode *node = root;memset(st, 0, sizeof(*st));st->top = -1;*returnSize = 0;if (root == NULL) {return array;}while (st->top > -1 || node != NULL) {while (node != NULL) {array[(*returnSize)++] = node->val;st->stk[++st->top] = node;node = node->left;}node = st->stk[st->top--];node = node->right;}printf("%d\n", *returnSize);return array;
}

145. 二叉树的后序遍历

解1: 递归

void postorder(struct TreeNode *root, int *ans, int *returnSize) {if (root != NULL) {postorder(root->left, ans, returnSize);postorder(root->right, ans, returnSize);ans[(*returnSize)++] = root->val;}
}int* postorderTraversal(struct TreeNode* root, int* returnSize) {int *ans = (int *)malloc(sizeof(int) * 100);*returnSize = 0;postorder(root, ans, returnSize);return ans;
}

解2: 迭代

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     struct TreeNode *left;*     struct TreeNode *right;* };*/
/*** Note: The returned array must be malloced, assume caller calls free().*/struct stack{int top;struct TreeNode *stk[100];
};void reverse(int *array, int n) {int left = 0, right = n-1;int tmp;while (left < right) {tmp = array[left];array[left] = array[right];array[right] = tmp;left++, right--;}
}int* postorderTraversal(struct TreeNode* root, int* returnSize) {int *array = (int *)malloc(sizeof(int) * 100);struct stack *st = (struct stack *)malloc(sizeof(*st));struct TreeNode *node = NULL;memset(st, 0, sizeof(*st));st->top = -1;*returnSize = 0;if (root == NULL) {return array;}st->stk[++st->top] = root;while (st->top > -1) {node = st->stk[st->top--];array[(*returnSize)++] = node->val;if (node->left) st->stk[++st->top] = node->left;if (node->right) st->stk[++st->top] = node->right;}reverse(array, *returnSize);return array;
}

94. 二叉树的中序遍历

解1: 递归

void inorder(struct TreeNode* root, int *array, int *idx) {if (root != NULL) {inorder(root->left, array, idx);array[(*idx)++] = root->val;inorder(root->right, array, idx);}
}int* inorderTraversal(struct TreeNode* root, int* returnSize) {int *array = (int *)malloc(sizeof(int) * 100);memset(array, 0, sizeof(int) * 100);*returnSize = 0;inorder(root, array, returnSize);return array;
}

解2: 迭代

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     struct TreeNode *left;*     struct TreeNode *right;* };*/
/*** Note: The returned array must be malloced, assume caller calls free().*/struct stack{int top;struct TreeNode *stk[100];
};int* inorderTraversal(struct TreeNode* root, int* returnSize) {int *array = (int *)malloc(sizeof(int) * 100);struct stack *st = (struct stack *)malloc(sizeof(*st));struct TreeNode *node = NULL;memset(st, 0, sizeof(*st));st->top = -1;*returnSize = 0;if (root == NULL) {return array;}node = root;while (st->top > -1 || node != NULL) {while (node != NULL) {st->stk[++st->top] = node;node = node->left;}node = st->stk[st->top--];array[(*returnSize)++] = node->val;node = node->right;}return array;
}