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剑指offer——面试题6:重建二叉树

2016-11-25 12:19 555 查看












BinaryTreeNode* Construct(int* preorder, int* inorder, int length)
{
if(preorder == NULL || inorder == NULL || length <= 0)
return NULL;

return ConstructCore(preorder, preorder + length - 1,
inorder, inorder + length - 1);
}

BinaryTreeNode* ConstructCore
(
int* startPreorder, int* endPreorder,
int* startInorder, int* endInorder
)
{
// 前序遍历序列的第一个数字是根结点的值
int rootValue = startPreorder[0];
BinaryTreeNode* root = new BinaryTreeNode();
root->m_nValue = rootValue;
root->m_pLeft = root->m_pRight = NULL;

if(startPreorder == endPreorder)
{
if(startInorder == endInorder && *startPreorder == *startInorder)
return root;
else
throw std::exception("Invalid input.");
}

// 在中序遍历中找到根结点的值
int* rootInorder = startInorder;
while(rootInorder <= endInorder && *rootInorder != rootValue)
++ rootInorder;

if(rootInorder == endInorder && *rootInorder != rootValue)
throw std::exception("Invalid input.");

int leftLength = rootInorder - startInorder;
int* leftPreorderEnd = startPreorder + leftLength;
if(leftLength > 0)
{
// 构建左子树
root->m_pLeft = ConstructCore(startPreorder + 1, leftPreorderEnd,
startInorder, rootInorder - 1);
}
if(leftLength < endPreorder - startPreorder)
{
// 构建右子树
root->m_pRight = ConstructCore(leftPreorderEnd + 1, endPreorder,
rootInorder + 1, endInorder);
}

return root;
}
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