105. Construct Binary Tree from Preorder and Inorder Traversal

• Difficulty: Medium

• Topics: Array, Tree, Depth-first Search

• Similar Questions:

  • Construct Binary Tree from Inorder and Postorder Traversal

Problem:

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

Solutions:

/**
 * 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) {
        return helper(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);
    }

    TreeNode* helper(vector<int>& preorder, int preorderLeft, int preorderRight, vector<int>& inorder, int inorderLeft, int inorderRight) {
        if (preorderRight < preorderLeft)   return NULL;
        TreeNode* root = new TreeNode(preorder[preorderLeft]);
        int pos = find(inorder.begin() + inorderLeft, inorder.begin() + inorderRight + 1, preorder[preorderLeft]) - inorder.begin();
        int leftLen = pos - inorderLeft;
        int rightLen = inorderRight - pos;
        root->left = helper(preorder, preorderLeft + 1, preorderLeft + 1 + leftLen - 1, inorder, inorderLeft, pos - 1);
        root->right = helper(preorder, preorderLeft + 1 + leftLen, preorderRight, inorder, pos + 1, inorderRight);

        return root;
    }
};