刷题114. Flatten Binary Tree to Linked List

一、题目说明

题目114. Flatten Binary Tree to Linked List，将一个二叉树“原地”压缩为“链表”形态的二叉树。难度为Medium！

二、我的解答

这个题目如果允许使用栈的话Easy，先序遍历二叉树，右子树入栈，左子树入栈。当栈不空的时候，将栈顶元素放到右子树即可。

class Solution{
	public:
		void flatten(TreeNode* root){
			
			//先根遍历 
			if(root==NULL) return;
			if(root->left==NULL && root->right==NULL) return;
			TreeNode *p,* cur = root;
			
			stack<TreeNode*> st;
			if(root->right !=NULL){
				st.push(root->right);
			}
			if(root->left !=NULL){
				st.push(root->left);
			}
			
			while(! st.empty()){
				p = st.top();
				st.pop();
				cur->left = NULL;
				cur->right = p;
				cur = cur->right;
				
				if(cur->right !=NULL){
					st.push(cur->right);
				}
				if(cur->left !=NULL){
					st.push(cur->left);
				}
			}
			
			return; 
		}
};

性能：

Runtime: 12 ms, faster than 27.18% of C++ online submissions for Flatten Binary Tree to Linked List.
Memory Usage: 11.6 MB, less than 8.33% of C++ online submissions for Flatten Binary Tree to Linked List.

三、优化措施

此处的“原地”，理解起来不能使用栈的。在这种情况下，将右子树作为左子树的最右下节点的右子树，左子树变为右子树即可。

class Solution{
	public:
		void flatten(TreeNode* root){
			if(root==NULL) return;
			if(root->left !=NULL){
				TreeNode* pre = root->left;
				while(pre->right !=NULL){
					pre = pre->right;
				}
				pre->right = root->right;
				root->right = root->left;
				root->left = NULL;
			}
			flatten(root->right);
			return; 
		}
};

性能如下：

Runtime: 4 ms, faster than 95.35% of C++ online submissions for Flatten Binary Tree to Linked List.
Memory Usage: 11.3 MB, less than 8.33% of C++ online submissions for Flatten Binary Tree to Linked List.

这个性能还一般，用morris方法（线索化二叉树），空间复杂度可以到O(1)。