leetcode solution for problem 105 : Construct Binary Tree from Preorder and Inorder Traversal

dsa-solutions/lc-solutions/0100-0199/0105-Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal.md

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+---
+id: Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal
+title: Construct Binary Tree from Preorder and Inorder Traversal
+sidebar_label: 0105-Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal
+tags:
+  - Array
+  - Hash Tabel
+  - Divide and Conquer
+  - Tree
+  - Binary Tree
+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."
+---
+
+
+### Problem 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.
+
+### Examples
+
+#### Example 1
+
+- **Input:** `preorder = [3,9,20,15,7], inorder=[9,3,15,20,7]`
+- **Output:** `[3,9,30,null,null,15,7]`
+
+#### Example 2
+
+- **Input:** `preorder = [-1], inorder=[-1]`
+- **Output:** `[-1]`
+
+### Constraints
+
+- $1 \leq \text{preorder.length} \leq 3000$
+- $\text{inorder.length} == \text{preorder.length}$
+- $-3000 \leq \text{preorder}[i], \text{inorder}[i] \leq 3000$
+- 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.
+
+### Solution Code
+
+#### Python
+
+```python
+class Solution:
+    def buildTree(self, P: List[int], I: List[int]) -> TreeNode:
+        M = {I[i]: i for i in range(len(I))}
+        return self.splitTree(P, M, 0, 0, len(P)-1)
+
+    def splitTree(self, P: List[int], M: dict, pix: int, ileft: int, iright: int) -> TreeNode:
+        rval = P[pix]
+        root, imid = TreeNode(rval), M[rval]
+        if imid > ileft:
+            root.left = self.splitTree(P, M, pix+1, ileft, imid-1)
+        if imid < iright:
+            root.right = self.splitTree(P, M, pix+imid-ileft+1, imid+1, iright)
+        return root
+```
+
+#### Java
+
+```java
+class Solution {
+    public TreeNode buildTree(int[] P, int[] I) {
+        Map<Integer, Integer> M = new HashMap<>();
+        for (int i = 0; i < I.length; i++)
+            M.put(I[i], i);
+        return splitTree(P, M, 0, 0, I.length-1);
+    }
+
+    private TreeNode splitTree(int[] P, Map<Integer, Integer> M, int pix, int ileft, int iright) {
+        int rval = P[pix], imid = M.get(rval);
+        TreeNode root = new TreeNode(rval);            
+        if (imid > ileft)
+            root.left = splitTree(P, M, pix+1, ileft, imid-1);
+        if (imid < iright)
+            root.right = splitTree(P, M, pix+imid-ileft+1, imid+1, iright);
+        return root;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    TreeNode* buildTree(vector<int>& P, vector<int>& I) {
+        unordered_map<int, int> M;
+        for (int i = 0; i < I.size(); i++)
+            M[I[i]] = i;
+        return splitTree(P, M, 0, 0, I.size()-1);
+    }
+
+private:
+    TreeNode* splitTree(vector<int>& P, unordered_map<int, int>& M, int pix, int ileft, int iright) {
+        int rval = P[pix], imid = M[rval];
+        TreeNode* root = new TreeNode(rval);            
+        if (imid > ileft)
+            root->left = splitTree(P, M, pix+1, ileft, imid-1);
+        if (imid < iright)
+            root->right = splitTree(P, M, pix+imid-ileft+1, imid+1, iright);
+        return root;
+    }
+};
+```
+#### Javascript
+
+```javascript
+var buildTree = function(P, I) {
+    let M = new Map()
+    for (let i = 0; i < I.length; i++)
+        M.set(I[i], i)
+    return splitTree(P, M, 0, 0, I.length-1)
+};
+
+var splitTree = function(P, M, pix, ileft, iright) {
+    let rval = P[pix],
+        root = new TreeNode(rval),
+        imid = M.get(rval)
+    if (imid > ileft)
+        root.left = splitTree(P, M, pix+1, ileft, imid-1)
+    if (imid < iright)
+        root.right = splitTree(P, M, pix+imid-ileft+1, imid+1, iright)
+    return root
+}
+```