LeetCode(105) Construct Binary Tree from Preorder and Inorder Traversal
题目
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
分析
给定一颗二叉树的前序和中序遍历序列,求该二叉树。
我们手动做过很多这样的题目,掌握了其规则~
前序遍历第一个元素为树的root节点,然后在中序序列中查找该值,元素左侧为左子树,右侧为右子树; 求出左子树个数count,在前序序列中 , 除去第一个节点,接下来的count个元素构成左子树的前序序列,其余的构成右子树的前序序列。
开始,没有采用迭代器,声明vector占用了大量空间,Memory Limit Exceeded。。。
代码为:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
if (preorder.empty() && inorder.empty())
return NULL;
//求树中节点个数
int size = preorder.size();
//先序遍历第一个节点为树的根节点
TreeNode *root = new TreeNode(preorder[0]);
int pos = 0;
//在中序遍历结果中查找根节点
for (int i=0; i<size; ++i)
{
if (inorder[i] == preorder[0])
{
pos = i;
break;
}//if
}//for
if (pos >= 0 && pos < size)
{
//则在inOrder中(0 , pos-1)为左子树中序遍历结果(pos+1,size-1)为右子树的中序遍历序列
//在preOrder中(1,pos)为左子树前序遍历结果(pos+1,size-1)为右子树前序遍历结果
vector<int> left_pre;
for (int j = 1; j <= pos; j++)
left_pre.push_back(preorder[j]);
vector<int> left_in;
for (int j = 0; j < pos; ++j)
left_in.push_back(inorder[j]);
root->left = buildTree(left_pre, left_in);
//构造右子树
vector<int> right_pre , right_in;
for (int j = pos + 1; j < size; j++)
{
right_pre.push_back(preorder[j]);
right_in.push_back(inorder[j]);
}
root->right = buildTree(right_pre, right_in);
}
return root;
}
};
然后,使用迭代器避免不必要的空间占用,AC~
AC代码
class Solution {
public:
template <typename Iter>
TreeNode* make(Iter pre_begin, Iter pre_end, Iter in_begin, Iter in_end) {
if (pre_begin == pre_end || in_begin == in_end)
return NULL;
//先序遍历第一个节点为树的根节点
TreeNode *root = new TreeNode(*pre_begin);
//在中序遍历结果中查找根节点
Iter iter = find(in_begin, in_end, *pre_begin);
int count = iter - in_begin;
if (iter != in_end)
{
//则在inOrder中(0 , pos-1)为左子树中序遍历结果(pos+1,size-1)为右子树的中序遍历序列
//在preOrder中(1,pos)为左子树前序遍历结果(pos+1,size-1)为右子树前序遍历结果
root->left = make(pre_begin + 1, pre_begin + count + 1, in_begin, iter);
//构造右子树
root->right = make(pre_begin + count + 1, pre_end, iter + 1, in_end);
}
return root;
}
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
if (preorder.empty() || inorder.empty())
return NULL;
return make(preorder.begin(), preorder.end(), inorder.begin(), inorder.end());
}
};
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