二叉树的深度遍历优先（前中后序遍历）

对于二叉树部分LeetCode还有很多简单但是好的题目：

LeetCode关于单值、翻转、对称、最大深度、另一颗树的子树二叉树题目解析链接: link.

LeetCode关于平衡二叉树和层序遍历解析链接: link.

1. LeetCode第144题—二叉树的前序遍历

链接: link.

/**
 * 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().
 */
 //因为你需要知道在开辟空间大小的时候，开辟多大才行，所以要算结点的个数
 int TreeSize(struct TreeNode* root)
 {
     
    if(root == NULL)
        return 0;
    else
        return 1+TreeSize(root->left) + TreeSize(root->right);
 }

//前序遍历
 void _preorderTraversal(struct TreeNode* root,int* array,int* pi)
 {
     
    if(root == NULL)
        return;
    array[(*pi)++] = root->val;
    _preorderTraversal(root->left,array,pi); //这里的pi是传过来的指针，可以直接的使用
    _preorderTraversal(root->right,array,pi);
 }

//这里值得注意的就是在传i的时候一定要传地址，因为只有传值才能做到在同一个i上++的效果
int* preorderTraversal(struct TreeNode* root, int* returnSize){
     
    int size = TreeSize(root);
    int* array = (int*)malloc(sizeof(int)*size);
    int i = 0;
    _preorderTraversal(root,array,&i); //思考题目就明白他需要你把结点的值都放在一个数组里面
    *returnSize = size;
    return array;
}

2. LeetCode第94题—二叉树的中序遍历

链接: link.

/**
 * 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().
 */

 int TreeSize(struct TreeNode* root)
 {
     
    if(root == NULL)
        return 0;
    else
        return 1+TreeSize(root->left)+TreeSize(root->right);
 }
 
//左子树 根 右子树
 void _inorderTraversal(struct TreeNode* root,int* array,int* pi)
 {
     
    if(root == NULL)
        return;
    _inorderTraversal(root->left,array,pi);
    array[(*pi)++] = root->val;
    _inorderTraversal(root->right,array,pi);
 }

int* inorderTraversal(struct TreeNode* root, int* returnSize){
     
    int size = TreeSize(root);
    int* array = (int*)malloc(sizeof(int)*size);
    int i = 0;
    _inorderTraversal(root,array,&i);
    *returnSize = size;
    return array;
}

3. LeetCode第145题—二叉树的后序遍历

链接: link.

/**
 * 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().
 */

  int TreeSize(struct TreeNode* root)
 {
     
    if(root == NULL)
        return 0;
    else
        return 1+TreeSize(root->left)+TreeSize(root->right);
 }

  void _postorderTraversal(struct TreeNode* root,int* array,int* pi)
 {
     
    if(root == NULL)
        return;
    _postorderTraversal(root->left,array,pi);
    _postorderTraversal(root->right,array,pi);
    array[(*pi)++] = root->val;
 }

int* postorderTraversal(struct TreeNode* root, int* returnSize){
     
    int size = TreeSize(root);
    int* array = (int*)malloc(sizeof(int)*size);
    int i = 0;
    _postorderTraversal(root,array,&i);
    *returnSize = size;
    return array;
}