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LeetCode OJ 94. Binary Tree Inorder Traversal

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) {
    int stack[1000], top = -1, i = 0;
    int *return_array = (int*)malloc(sizeof(int) * 1000);
    
    while(-1 != 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;
}

 

LeetCode OJ 94. Binary Tree Inorder Traversal