106. Construct Binary Tree from Inorder and Postorder Traversal

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106. Construct Binary Tree from Inorder and Postorder Traversal

题目

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

解析

// 106. Construct Binary Tree from Inorder and Postorder Traversal
class Solution_106 {
public:
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) { //这样消耗内存多些

if (inorder.size()==0||postorder.size()==0||inorder.size()!=postorder.size())
{
return NULL;
}

int len = postorder.size();
TreeNode* root = new TreeNode(postorder[len-1]);

//bug:
//terminate called after throwing an instance of 'std::bad_alloc'
//what() : std::bad_alloc

auto pos = find(inorder.begin(), inorder.end(), postorder[len - 1]);
vector<int> inorder_l(inorder.begin(),pos);
vector<int> inorder_r(pos + 1, inorder.end());
vector<int> postorder_l(postorder.begin(), postorder.begin() + inorder_l.size());
vector<int> postorder_r(postorder.begin()+inorder_l.size(),postorder.end()-1);

if (inorder_l.size()>0)
{
root->left = buildTree(inorder_l, postorder_l);
}

if (inorder_r.size()>0)
{
root->right = buildTree(inorder_r, postorder_r);
}

return root;
}

public:
TreeNode* buildTreeHelper(vector<int>& inorder, int l1, int r1, vector<int>& postorder, int l2, int r2) //在原数组上操作，不需要额外空间
{
if (l1>r1||l2>r2)
{
return NULL;
}

TreeNode* root = new TreeNode(postorder[r2]);
int i = 0;
for ( i= l1; i <= r1;i++) // for ( i= 0; i < inorder.size();i++) //递归实现参数要能进入下次递归
{
if (inorder[i]==postorder[r2])
{
break;
}
}

root->left = buildTreeHelper(inorder,l1,i-1,postorder,l2,l2+(i-1-l1)); //慢慢体会下标的准确性
root->right = buildTreeHelper(inorder, i + 1, r1, postorder, l2 + (i - l1), r2-1);

return root;
}

TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder)
{
if (inorder.size()==0||postorder.size()==0||inorder.size()!=postorder.size())
{
return NULL;
}

return buildTreeHelper(inorder, 0, inorder.size() - 1, postorder,0, postorder.size() - 1);
}
};

题目来源