Construct Binary Tree from Preorder and Inorder Traversal Problem


Description

LeetCode Problem 105.

Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

Example 1:

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Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

Example 2:

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Input: preorder = [-1], inorder = [-1]
Output: [-1]

Constraints:

  • 1 <= preorder.length <= 3000
  • inorder.length == preorder.length
  • -3000 <= preorder[i], inorder[i] <= 3000
  • preorder and inorder consist of unique values.
  • Each value of inorder also appears in preorder.
  • preorder is guaranteed to be the preorder traversal of the tree.
  • inorder is guaranteed to be the inorder traversal of the tree.


Sample C++ Code

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class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if (preorder.size() == 0)
            return NULL;
        TreeNode* root = new TreeNode(preorder[0]);
        
        vector<int> left_pre, right_pre, left_in, right_in;
        bool isleft = true;
        for (int i = 0; i < inorder.size(); i ++) {
            if (inorder[i] == preorder[0]) {
                isleft = false;
                continue;
            }
            if (isleft) {
                left_pre.push_back(preorder[i+1]);
                left_in.push_back(inorder[i]);
            } else {
                right_pre.push_back(preorder[i]);
                right_in.push_back(inorder[i]);
            }
        }
        
        root->left = buildTree(left_pre, left_in);
        root->right = buildTree(right_pre, right_in);
        
        return root;
    }
};




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