Construct Binary Tree from Preorder and Inorder Traversal leetcode

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

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

 

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伪代码：

buildTree(pbeg, ibeg, iend)
    front = preorder[pbeg]
    root  = new Node(front)
    find front idx from inorder[ibeg to iend]
    root->left  = buildTree(pbeg+1,            ibeg,  idx)
    root->right = buildTree(pbeg+1+index-ibeg, idx+1, iend)
    return root

递归方法实现：

TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder, size_t preBegin, size_t inBegin, size_t inEnd)
{
    if (inBegin == inEnd)
        return nullptr;
    int front = preorder[preBegin];
    TreeNode* root = new TreeNode(front);
    int index = inBegin;
    while (index < inEnd && inorder[index] != front)
        index++;
    root->left = buildTree(preorder, inorder, preBegin + 1, inBegin, index);
    root->right = buildTree(preorder, inorder, preBegin + 1 + index - inBegin, index+1, inEnd);
    return root;
}

TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
    return buildTree(preorder, inorder, 0, 0, inorder.size());
}