NO.107 二叉树的层次遍历II

给定一个二叉树，返回其节点值自底向上的层次遍历。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

例如：
给定二叉树 [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

返回其自底向上的层次遍历为：

[
  [15,7],
  [9,20],
  [3]
]

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */


/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
void DFS(struct TreeNode* root,int* returnSize,int** returnColumnSizes,int target_size,int ***ret)
{
    if(!root)return;
    if(*returnSize<target_size)
    {
        *returnColumnSizes=(int *)realloc(*returnColumnSizes,sizeof(int )*target_size);
        memmove((*returnColumnSizes)+1,*returnColumnSizes,sizeof(int)*(* returnSize));
        (*returnColumnSizes)[0]=0;
        *ret=(int **)realloc(*ret,sizeof(int *)*target_size);
        memmove((*ret)+1,*ret,sizeof(int*)*(* returnSize));
        (* ret)[0]=NULL;
        (* returnSize)++;
    }
    (*returnColumnSizes)[(* returnSize)-target_size]++;
    (* ret)[(* returnSize)-target_size]=(int *)realloc((* ret)[(* returnSize)-target_size],sizeof(int)*(*returnColumnSizes)[(* returnSize)-target_size]);
    (* ret)[(* returnSize)-target_size][(*returnColumnSizes)[(* returnSize)-target_size]-1]=root->val;
    DFS(root->left,returnSize,returnColumnSizes,target_size+1,ret);
    DFS(root->right,returnSize,returnColumnSizes,target_size+1,ret);
}

int** levelOrderBottom(struct TreeNode* root, int* returnSize, int** returnColumnSizes){
    int **ret=NULL;
    * returnSize=0;
    * returnColumnSizes=NULL;
    DFS(root,returnSize,returnColumnSizes,1,&ret);
    return ret;
}

执行用时 :20 ms, 在所有C提交中击败了57.35% 的用户

内存消耗 :17.7 MB, 在所有C提交中击败了12.12%的用户