leetcode_894. All Possible Full Binary Trees

https://leetcode.com/problems/all-possible-full-binary-trees/

给定节点个数，求所有可能二叉树，该二叉树所有节点要么有0个子节点要么有两个子节点。返回所有二叉树的头指针。

一开始一直想的是从根节点开始建树，一直想不出来方法。后来想到可以从子节点开始建树，问题就很好解决了。

class Solution
{
public:vector<TreeNode*> allPossibleFBT(int N){vector<TreeNode*> ret;if(N == 1){TreeNode* rt = new TreeNode(0);ret.push_back(rt);return ret;}for(int i=1; i<=(N-1)/2; i+=2)  //左子树的节点数{vector<TreeNode*> left = allPossibleFBT(i);  　　  //创建所有可能左子树vector<TreeNode*> right = allPossibleFBT(N-1-i);  //创建所有可能的右子树for(int j=0;j<left.size();j++)  　　　　//遍历所有左子树for(int k=0;k<right.size();k++)    //遍历所有右子树{TreeNode * rt = new TreeNode(0);   //创建根节点rt->left = left[j];rt->right = right[k];ret.push_back(rt);if(i != N-1-i)   　　　//如果左右子树节点数不同，交换左右子树也是一种可能{TreeNode * rt2 = new TreeNode(0);rt2->left = right[k];rt2->right = left[j];ret.push_back(rt2);}}}return ret;}
};

转载于:https://www.cnblogs.com/jasonlixuetao/p/10582693.html

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