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)。
原文地址:https://www.cnblogs.com/siweihz/p/12267743.html
时间: 2024-07-30 13:45:32