码迷,mamicode.com
首页 > 其他好文 > 详细

LeetCode OJ 144. Binary Tree Preorder Traversal

时间:2016-11-07 09:44:09      阅读:176      评论:0      收藏:0      [点我收藏+]

标签:returns   val   ++   his   log   scribe   roo   node   com   

Given a binary tree, return the preorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

 

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

 

Subscribe to see which companies asked this question

解答

注意空树的情况……

/**
 * 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* preorderTraversal(struct TreeNode* root, int* returnSize) {
    int top = -1, i = 0;
    int *return_array = (int*)malloc(sizeof(int) * 1000);
    struct TreeNode *stack[1000];
    
    if(NULL == root){
        *returnSize = i;
        return return_array;
    }
    stack[++top] = root;
    while(-1 != top){
        root = stack[top--];
        return_array[i++] = root->val;
        if(NULL != root->right)
            stack[++top] = root->right;
        if(NULL != root->left)
            stack[++top] = root->left;
    }
    *returnSize = i;
    return return_array;
}

 

LeetCode OJ 144. Binary Tree Preorder Traversal

标签:returns   val   ++   his   log   scribe   roo   node   com   

原文地址:http://www.cnblogs.com/YuNanlong/p/6033092.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!