added leetcode problem 106 solution

dsa-solutions/lc-solutions/0100-0199/0106-construct-binary-tree-from-inorder.md

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+---
+id: Construct-Binary-Tree-from-Inorder-and-postorder-Traversal
+title: Construct Binary Tree from Inorder and Postorder Traversal
+sidebar_label: 0106-Construct-Binary-Tree-from-Inorder-and-postorder-Traversal
+tags:
+  - Array
+  - Hash Tabel
+  - Divide and Conquer
+  - Tree
+  - Binary Tree
+description: "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.
+
+ "
+---
+
+
+### Problem Description
+
+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
+
+- $1 \leq \text{inorder.length} \leq 3000$
+- $\text{postorder.length} == \text{inorder.length}$
+- $-3000 \leq \text{inorder}[i], \text{postorder}[i] \leq 3000$
+- postorder and inorder consist of unique values.
+- Each value of postorder also appears in inorder.
+- postorder is guaranteed to be the postorder traversal of the tree.
+- inorder is guaranteed to be the inorder traversal of the tree.
+
+### Solution Code
+
+#### Python
+
+```python
+class Solution:
+    def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
+        n=len(inorder)
+        poststart=instart=0
+        postend=inend=n-1
+        d={}
+        for i in range(n):
+            d[inorder[i]]=i
+        return self.constructTree(postorder,poststart,postend,inorder,instart,inend,d)
+
+    def constructTree(self,postorder,poststart,postend,inorder,instart,inend,d):
+        if poststart>postend or instart>inend:
+            return None
+        root=TreeNode(postorder[postend])
+        elem=d[root.val]
+        nelem=elem-instart
+        root.left=self.constructTree(postorder,poststart,poststart+nelem-1,inorder,instart,elem-1,d)
+        root.right=self.constructTree(postorder,poststart+nelem,postend-1,inorder,elem+1,inend,d)
+        return root
+```
+
+#### Java
+
+```java
+class Solution 
+{
+    public TreeNode buildTree(int[] inorder, int[] postorder) {
+    // If either of the input arrays are empty, the tree is empty, so return null
+    if (inorder.length == 0 || postorder.length == 0) return null;
+
+    // Initialize indices to the last elements of the inorder and postorder traversals
+    int ip = inorder.length - 1;
+    int pp = postorder.length - 1;
+
+    // Create an empty stack to help us build the binary tree
+    Stack<TreeNode> stack = new Stack<TreeNode>();
+    // Initialize prev to null since we haven't processed any nodes yet
+    TreeNode prev = null;
+    // Create the root node using the last element in the postorder traversal
+    TreeNode root = new TreeNode(postorder[pp]);
+    // Push the root onto the stack and move to the next element in the postorder traversal
+    stack.push(root);
+    pp--;
+
+    // Process the rest of the nodes in the postorder traversal
+    while (pp >= 0) {
+        // While the stack is not empty and the top of the stack is the current inorder element
+        while (!stack.isEmpty() && stack.peek().val == inorder[ip]) {
+            // The top of the stack is the parent of the current node, so pop it off the stack and update prev
+            prev = stack.pop();
+            ip--;
+        }
+        // Create a new node for the current postorder element
+        TreeNode newNode = new TreeNode(postorder[pp]);
+        // If prev is not null, the parent of the current node is prev, so attach the node as the left child of prev
+        if (prev != null) {
+            prev.left = newNode;
+        // If prev is null, the parent of the current node is the current top of the stack, so attach the node as the right child of the current top of the stack
+        } else if (!stack.isEmpty()) {
+            TreeNode currTop = stack.peek();
+            currTop.right = newNode;
+        }
+        // Push the new node onto the stack, reset prev to null, and move to the next element in the postorder traversal
+        stack.push(newNode);
+        prev = null;
+        pp--;
+    }
+
+    // Return the root of the binary tree
+    return root;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
+    // If either of the input vectors are empty, the tree is empty, so return null
+    if (inorder.size() == 0 || postorder.size() == 0) return nullptr;
+
+    // Initialize indices to the last elements of the inorder and postorder traversals
+    int ip = inorder.size() - 1;
+    int pp = postorder.size() - 1;
+
+    // Create an empty stack to help us build the binary tree
+    stack<TreeNode*> st;
+    // Initialize prev to null since we haven't processed any nodes yet
+    TreeNode* prev = nullptr;
+    // Create the root node using the last element in the postorder traversal
+    TreeNode* root = new TreeNode(postorder[pp]);
+    // Push the root onto the stack and move to the next element in the postorder traversal
+    st.push(root);
+    pp--;
+
+    // Process the rest of the nodes in the postorder traversal
+    while (pp >= 0) {
+        // While the stack is not empty and the top of the stack is the current inorder element
+        while (!st.empty() && st.top()->val == inorder[ip]) {
+            // The top of the stack is the parent of the current node, so pop it off the stack and update prev
+            prev = st.top();
+            st.pop();
+            ip--;
+        }
+        // Create a new node for the current postorder element
+        TreeNode* newNode = new TreeNode(postorder[pp]);
+        // If prev is not null, the parent of the current node is prev, so attach the node as the left child of prev
+        if (prev != nullptr) {
+            prev->left = newNode;
+        // If prev is null, the parent of the current node is the current top of the stack, so attach the node as the right child of the current top of the stack
+        } else if (!st.empty()) {
+            TreeNode* currTop = st.top();
+            currTop->right = newNode;
+        }
+        // Push the new node onto the stack, reset prev to null, and move to the next element in the postorder traversal
+        st.push(newNode);
+        prev = nullptr;
+        pp--;
+    }
+
+    // Return the root of the binary tree
+    return root;
+    }
+};
+```
+#### Javascript
+
+```javascript
+var buildTree = function(inorder, postorder) {
+    if (inorder.length == 0 || postorder.length == 0) {
+        return null;
+    }
+    var rootVal = postorder[postorder.length - 1];
+    var root = new TreeNode(rootVal);
+    var rootIndex = inorder.indexOf(rootVal);
+    var leftInorder = inorder.slice(0, rootIndex);
+    var rightInorder = inorder.slice(rootIndex + 1);
+    var leftPostorder = postorder.slice(0, leftInorder.length);
+    var rightPostorder = postorder.slice(leftInorder.length, postorder.length - 1);
+    root.left = buildTree(leftInorder, leftPostorder);
+    root.right = buildTree(rightInorder, rightPostorder);
+    return root;
+};
+```