Construct Binary Tree from Inorder and Postorder Traversal
Problem Statement - link #
Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.
Examples #
Example 1:
Input: inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
Output: [3,9,20,null,null,15,7]
Example 2:
Input: inorder = [-1], postorder = [-1]
Output: [-1]
Constraints #
1 <= inorder.length <= 3000postorder.length == inorder.length-3000 <= inorder[i], postorder[i] <= 3000inorderandpostorderconsist of unique values.- Each value of
postorderalso appears in inorder. inorderis guaranteed to be the inorder traversal of the tree.postorderis guaranteed to be the postorder traversal of the tree.
Solutions #
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int xi;
vector<int> in, post;
TreeNode* rc(int l, int r){
if(l>r) return NULL;
int i = 0;
while(in[i]!=post[xi]) i++;
xi--;
TreeNode* node = new TreeNode(in[i]);
node->right = rc(i+1,r);
node->left = rc(l,i-1);
return node;
}
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
in = inorder;
post = postorder;
xi = post.size()-1;
return rc(0, in.size()-1);
}
};