105. Construct Binary Tree from Preorder and Inorder Traversal

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

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

思路

c++11

/**
 * 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:
    using cIter = vector<int>::const_iterator;
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        return buildTree(preorder.cbegin(), preorder.cend(), inorder.cbegin(), inorder.cend());
    }
    TreeNode *buildTree(cIter preBeg, cIter preEnd, cIter inBeg, cIter inEnd) {
        if (preBeg >= preEnd || inBeg >= inEnd) return NULL;
        TreeNode *root = new TreeNode(*preBeg);
        auto inRoot = std::find(inBeg, inEnd, root->val);
        root->left = buildTree(next(preBeg), next(preBeg, inRoot-inBeg+1), inBeg, inRoot);
        root->right = buildTree(next(preBeg, inRoot-inBeg+1), preEnd, next(inRoot), inEnd);
        return root;
    }
};