LeetCode 1008 - Construct Binary Search Tree from Preorder Traversal

Construct Binary Search Tree from a given Preorder

Construct a Binary Search Tree from given postorder

https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal/discuss/252232/JavaC%2B%2BPython-O(N)-Solution

Find the left part and right part,
then recursively construct the tree.

Solution 1:

Time: O(HlogN)

    def bstFromPreorder(self, A):
        if not A: return None
        root = TreeNode(A[0])
        i = bisect.bisect(A, A[0])
        root.left = self.bstFromPreorder(A[1:i])
        root.right = self.bstFromPreorder(A[i:])
        return root

Solution 2

Give the function a bound the maximum number it will handle.
The left recursion will take the elements smaller than node.val
The right recursion will take the remaining elements smaller than bound

Time: O(N)

    int i = 0;
    public TreeNode bstFromPreorder(int[] A) {
        return bstFromPreorder(A, Integer.MAX_VALUE);
    }

    public TreeNode bstFromPreorder(int[] A, int bound) {
        if (i == A.length || A[i] > bound) return null;
        TreeNode root = new TreeNode(A[i++]);
        root.left = bstFromPreorder(A, root.val);
        root.right = bstFromPreorder(A, bound);
        return root;
    }

X. https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal/discuss/252754/Java-Stack-Iterative-Solution

    public TreeNode bstFromPreorder(int[] preorder) {
        if (preorder == null || preorder.length == 0) {
            return null;
        }
        Stack<TreeNode> stack = new Stack<>();
        TreeNode root = new TreeNode(preorder[0]);
        stack.push(root);
        for (int i = 1; i < preorder.length; i++) {
            TreeNode node = new TreeNode(preorder[i]);
            if (preorder[i] < stack.peek().val) {                
                stack.peek().left = node;                
            } else {
                TreeNode parent = stack.peek();
                while (!stack.isEmpty() && preorder[i] > stack.peek().val) {
                    parent = stack.pop();
                }
                parent.right = node;
            }
            stack.push(node);            
        }
        return root;
    }

http://www.noteanddata.com/leetcode-1008-Construct-Binary-Search-Tree-from-Preorder-Traversal-java-solution-note.html

平均O(N*lgN)，最差O(N^2), 相当于对于树的每一层，都对N个元素做了一次访问，树高平均是lgN, 最差是N

public TreeNode bstFromPreorder(int[] preorder) {
    return tree(preorder, 0, preorder.length-1);
}

public TreeNode tree(int[] preorder, int from, int to) {
    if(from > to) return null;
    TreeNode root = new TreeNode(preorder[from]);
    boolean found = false;
    for(int i = from+1; i <= to; ++i) {
        if(preorder[i] > preorder[from]) {
            root.left = tree(preorder, from+1, i-1);
            root.right = tree(preorder, i, to);
            found = true;
            break;
        }
    }
    if(!found) {
        root.left = tree(preorder, from+1, to);
    }
    
    return root;
}

https://www.ideserve.co.in/learn/serialize-deserialize-binary-search-tree-using-post-order-traversal

    public TreeNode deserializeArrayOptimized(int[] postorder, int[] currIndex, int min, int max)

{

        if (currIndex[0] < 0) return null;

        TreeNode root = null;

        if ((postorder[currIndex[0]] > min) && (postorder[currIndex[0]] < max))

{

            root = new TreeNode(postorder[currIndex[0]]);

            currIndex[0] -= 1;

            root.right = deserializeArrayOptimized(postorder, currIndex, root.val, max);

            root.left = deserializeArrayOptimized(postorder, currIndex, min, root.val);

}

        return root;

}

    private int findDivision(int[] postorder, int value, int low, int high)

{

        int i = high;

        for (; i >= low; i--)

{

            if (value > postorder[i])

                break;

}

        return i;

}

    public TreeNode deserializeArray(int[] postorder, int low, int high)

{

        if (low > high) return null;

        TreeNode root = new TreeNode(postorder[high]);

        int divIndex = findDivision(postorder, root.val, low, high - 1);

        root.left = deserializeArray(postorder, low, divIndex);

        root.right = deserializeArray(postorder, divIndex + 1, high - 1);

        return root;

}

Construct a Binary Search Tree from given postorder - GeeksforGeeks
Related: Construct BST from given preorder traversal
Given postorder traversal of a binary search tree, construct the BST.
For example, if the given traversal is {1, 7, 5, 50, 40, 10}, then following tree should be constructed and root of the tree should be returned.

     10
   /   \
  5     40
 /  \      \
1    7      50

Method 2 ( O(n) time complexity )
The trick is to set a range {min .. max} for every node. Initialize the range as {INT_MIN .. INT_MAX}. The last node will definitely be in range, so create root node. To construct the left subtree, set the range as {INT_MIN …root->data}. If a values is in the range {INT_MIN .. root->data}, the values is part part of left subtree. To construct the right subtree, set the range as {root->data .. INT_MAX}.

class Node  

{

    int data; 

    Node left, right; 

    Node(int data)  

{

        this.data = data; 

        left = right = null; 

}

}

// Class containing variable that keeps a track of overall 

// calculated postindex 

class Index  

{

    int postindex = 0; 

}

class BinaryTree  

{

    // A recursive function to construct BST from post[].  

    // postIndex is used to keep track of index in post[]. 

    Node constructTreeUtil(int post[], Index postIndex, 

            int key, int min, int max, int size)  

{

        // Base case 

        if (postIndex.postindex < 0) 

            return null; 

        Node root = null; 

        // If current element of post[] is in range, then 

        // only it is part of current subtree 

        if (key > min && key < max)  

{

            // Allocate memory for root of this subtree and decrement 

            // *postIndex 

            root = new Node(key); 

            postIndex.postindex = postIndex.postindex - 1; 

            if (postIndex.postindex > 0)  

{

                // All nodes which are in range {key..max} will go in  

                // right subtree, and first such node will be root of right 

                // subtree 

                root.right = constructTreeUtil(post, postIndex,  

                        post[postIndex.postindex],key, max, size); 

                // Contruct the subtree under root 

                // All nodes which are in range {min .. key} will go in left 

                // subtree, and first such node will be root of left subtree. 

                root.left = constructTreeUtil(post, postIndex,  

                        post[postIndex.postindex],min, key, size); 

}

}

        return root; 

}

    // The main function to construct BST from given postorder 

    // traversal. This function mainly uses constructTreeUtil() 

    Node constructTree(int post[], int size)  

{

        Index index = new Index(); 

        index.postindex = size - 1; 

        return constructTreeUtil(post, index, post[index.postindex], 

                Integer.MIN_VALUE, Integer.MAX_VALUE, size); 

}

https://www.techiedelight.com/build-binary-search-tree-from-postorder-sequence/

Node* buildTree(vector<int> const &postorder, int& pIndex,

                int min, int max)

{

    // Base case

    if (pIndex < 0)

        return nullptr;

    // Return if next element of postorder traversal from the end

    // is not in valid range

    int curr = postorder[pIndex];

    if (curr < min || curr > max)

        return nullptr;

    // Construct the root node and decrement pIndex

    Node* root = newNode(curr);

    pIndex--;

    // Construct left and right subtree of the root node.

    // Build right subtree before left subtree since the values are

    // being read from the end of the postorder sequence

    // Since all elements in the right subtree of a BST must be greater

    // than the value of root node, set range as [curr+1..max] and recur

    root->right = buildTree(postorder, pIndex, curr+1, max);

    // Since all elements in the left subtree of a BST must be less

    // than the value of root node, set range as [min, curr-1] and recur

    root->left = buildTree(postorder, pIndex, min, curr-1);

    return root;

}

// Build a binary search tree from its postorder sequence

Node* buildTree(vector<int> const &postorder)

{

    // start from the root node (last element in postorder sequence)

    int postIndex = postorder.size() - 1;

    // set range of the root node as [INT_MIN, INT_MAX] and recur

    return buildTree(postorder, postIndex, INT_MIN, INT_MAX);

}

X. https://www.geeksforgeeks.org/construct-a-bst-from-given-postorder-traversal-using-stack/

Push root of the BST to the stack i.e, last element of the array.
Start traversing the array in reverse, if next element is > the element at the top of the stack then,
set this element as the right child of the element at the top of the stack and also push it to the stack.
Else if, next element is < the element at the top of the stack then,
start popping all the elements from the stack until either the stack is empty or the current element becomes > the element at the top of the stack.
Make this element left child of the last popped node and repeat the above steps until the array is traversed completely.

    Node constructTreeUtil(int post[], int n) 

    { 

        // Last node is root 

        Node root = new Node(post[n - 1]); 

        Stack<Node> s = new Stack<>(); 

        s.push(root); 

        // Traverse from second last node 

        for (int i = n - 2; i >= 0; --i) { 

            Node x = new Node(post[i]); 

            // Keep popping nodes while top() of stack 

            // is greater. 

            Node temp = null; 

            while (!s.isEmpty() && post[i] < s.peek().data)  

                temp = s.pop();       

            // Make x as left child of temp    

            if (temp != null)  

                temp.left = x;       

            // Else make x as right of top       

            else

                s.peek().right = x; 

            s.push(x); 

        } 

        return root; 

    } 

    // Function that calls the method which contructs the tree 

    Node constructTree(int post[], int size) 

    { 

        return constructTreeUtil(post, size); 

    }

Method 1 ( O(n^2) time complexity )

The last element of postorder traversal is always root. We first construct the root. Then we find the index of last element which is smaller than root. Let the index be ‘i’. The values between 0 and ‘i’ are part of left subtree, and the values between ‘i+1′ and ‘n-2′ are part of right subtree. Divide given post[] at index “i” and recur for left and right sub-trees.

For example in {1, 7, 5, 40, 50, 10}, 10 is the last element, so we make it root. Now we look for the last element smaller than 10, we find 5. So we know the structure of BST is as following.

             10
           /    \
          /      \
  {1, 7, 5}       {50, 40}

We recursively follow above steps for subarrays {1, 7, 5} and {40, 50}, and get the complete tree.

https://www.techiedelight.com/build-binary-search-tree-from-postorder-sequence/

struct Node* constructBST(int postorder[], int start, int end)

{

    // base case

    if (start > end)

        return NULL;

    // Construct the root node of the subtree formed by keys of the

    // postorder sequence in range [start, end]

    struct Node *node = newNode(postorder[end]);

    // search the index of last element in current range of postorder

    // sequence which is smaller than the value of root node

    int i;

    for (i = end; i >=start; i--) {

        if (postorder[i] < node->key)

            break;

}

    // Build right subtree before left subtree since the values are

    // being read from the end of the postorder sequence

    // recursively construct the right subtree

    node->right = constructBST(postorder, i + 1, end - 1);

    // recursively construct the left subtree

    node->left = constructBST(postorder, start, i);

    // return current node

    return node;

}

http://stackoverflow.com/questions/21173541/construct-binary-search-tree-from-post-order-traversal-in-java

public static Node buildBinarytreefromPostOrder(int[] post, int start, int end)
{
    if (end < start)
        return null;

    Node root = new Node(post[end]);

    if (end == start)
        return root;

    int i;
    for (i = end; i >= start; i--)
        if (post[i] < root.data)
            break;

    root.left = buildBinarytreefromPostOrder(post, start, i);
    root.right = buildBinarytreefromPostOrder(post, i + 1, end - 1);

    return root;
}

Read full article from Construct a Binary Search Tree from given postorder - GeeksforGeeks

LeetCode 1008 - Construct Binary Search Tree from Preorder Traversal

Solution 1:

Solution 2

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