Construct Binary Tree from Preorder and Inorder Traversal

Description

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:

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

Example 2:

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.

Code

rootIdx++; 就是一路按照 preorder 的 traversal 順序。注意 rootIdx 是 pass by reference！

class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int rootIdx = 0;
        return build(preorder, inorder, rootIdx, 0, inorder.size()-1);
    }
    
    TreeNode* build(vector<int>& preorder, vector<int>& inorder, int& rootIdx, int left, int right) {
        if (left > right) return NULL;
        int pivot = left;  // find the root from inorder
        while(inorder[pivot] != preorder[rootIdx]) pivot++;
        
        rootIdx++;
        TreeNode* newNode = new TreeNode(inorder[pivot]);
        newNode->left = build(preorder, inorder, rootIdx, left, pivot-1);
        newNode->right = build(preorder, inorder, rootIdx, pivot+1, right);
        return newNode;
    }
};

Source

• Construct Binary Tree from Preorder and Inorder Traversal - LeetCode