Construct Binary Tree from Inorder and Postorder Traversal LeetCode Solution

Problem – Construct Binary Tree from Inorder and Postorder Traversal LeetCode Solution

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.

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

Construct Binary Tree from Inorder and Postorder Traversal LeetCode Solution in Python

class Solution:
    def buildTree(self, inorder, postorder):
        if not inorder or not postorder:
            return None
        
        root = TreeNode(postorder.pop())
        inorderIndex = inorder.index(root.val) # Line A

        root.right = self.buildTree(inorder[inorderIndex+1:], postorder) # Line B
        root.left = self.buildTree(inorder[:inorderIndex], postorder) # Line C

        return root

Construct Binary Tree from Inorder and Postorder Traversal LeetCode Solution in C++

class Solution {
public:
    TreeNode *Tree(vector<int>& in, int x, int y,vector<int>& po,int a,int b){
        if(x > y || a > b)return nullptr;
        TreeNode *node = new TreeNode(po[b]);
        int SI = x;  
        while(node->val != in[SI])SI++;
        node->left  = Tree(in,x,SI-1,po,a,a+SI-x-1);
        node->right = Tree(in,SI+1,y,po,a+SI-x,b-1);
        return node;
    }
    TreeNode* buildTree(vector<int>& in, vector<int>& po){
        return Tree(in,0,in.size()-1,po,0,po.size()-1);
    }
};

Construct Binary Tree from Inorder and Postorder Traversal LeetCode Solution in Java

public TreeNode buildTreePostIn(int[] inorder, int[] postorder) {
	if (inorder == null || postorder == null || inorder.length != postorder.length)
		return null;
	HashMap<Integer, Integer> hm = new HashMap<Integer,Integer>();
	for (int i=0;i<inorder.length;++i)
		hm.put(inorder[i], i);
	return buildTreePostIn(inorder, 0, inorder.length-1, postorder, 0, 
                          postorder.length-1,hm);
}

private TreeNode buildTreePostIn(int[] inorder, int is, int ie, int[] postorder, int ps, int pe, 
                                 HashMap<Integer,Integer> hm){
	if (ps>pe || is>ie) return null;
	TreeNode root = new TreeNode(postorder[pe]);
	int ri = hm.get(postorder[pe]);
	TreeNode leftchild = buildTreePostIn(inorder, is, ri-1, postorder, ps, ps+ri-is-1, hm);
	TreeNode rightchild = buildTreePostIn(inorder,ri+1, ie, postorder, ps+ri-is, pe-1, hm);
	root.left = leftchild;
	root.right = rightchild;
	return root;
}

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