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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
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