106. Construct Binary Tree from Inorder and Postorder Travers

My solution Time limit exceed

// Time Limit Exceedpublic static TreeNode buildTree(int[] inorder, int[] postorder) {Map<Integer, Integer> map = new HashMap<Integer, Integer>();for (int i = 0; i < inorder.length; i++) {map.put(inorder[i], i);}return helper(postorder.length - 1, 0, inorder.length - 1, postorder, inorder, map);}public static TreeNode helper(int postEnd, int inStart, int inEnd, int[] postorder, int[] inorder,Map<Integer, Integer> map) {if (postEnd < 0 || inStart > inEnd) {return null;}TreeNode root = new TreeNode(postorder[postEnd]);int index = map.get(root.val);root.right = helper(postEnd - 1, index + 1, inEnd, postorder, inorder, map);root.left = helper(postEnd - index + inStart, inStart, index - 1, postorder, inorder, map);return root;}

Solution 2

public static TreeNode buildTree2(int[] inorder, int[] postorder) {if (inorder == null || postorder == null || inorder.length != postorder.length)return null;HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();for (int i = 0; i < inorder.length; ++i) {map.put(inorder[i], i);}return helper2(0, inorder.length - 1, 0, postorder.length - 1, inorder, postorder, map);}private static TreeNode helper2(int inStart, int inEnd, int postStart, int postEnd,  int[] inorder, int[] postorder,HashMap<Integer, Integer> map) {if (postStart > postEnd || inStart > inEnd)return null;TreeNode root = new TreeNode(postorder[postEnd]);int index = map.get(postorder[postEnd]);TreeNode leftchild = helper2(inStart, index - 1, postStart, postStart + index - inStart - 1, inorder, postorder, map);TreeNode rightchild = helper2(index + 1, inEnd, postStart + index - inStart, postEnd - 1, inorder, postorder, map);root.left = leftchild;root.right = rightchild;return root;}

Solution 3

//https://leetcode.com/discuss/10961/my-recursive-java-code-with-o-n-time-and-o-n-spaceint pInorder;   // index of inorder arrayint pPostorder; // index of postorder arrayprivate TreeNode helper3(int[] inorder, int[] postorder, TreeNode end) {    if (pPostorder < 0) {        return null;    }    // create root node    TreeNode n = new TreeNode(postorder[pPostorder--]);    // if right node exist, create right subtree    if (inorder[pInorder] != n.val) {        n.right = helper3(inorder, postorder, n);    }    pInorder--;    // if left node exist, create left subtree    if ((end == null) || (inorder[pInorder] != end.val)) {        n.left = helper3(inorder, postorder, end);    }    return n;}public TreeNode buildTree3(int[] inorder, int[] postorder) {    pInorder = inorder.length - 1;    pPostorder = postorder.length - 1;    return helper3(inorder, postorder, null);}

Solution 4

//https://leetcode.com/discuss/15115/my-comprehension-of-o-n-solution-from-%40hongzhi public TreeNode buildTree4(int[] inorder, int[] postorder) {        if (inorder == null || inorder.length < 1) return null;        int i = inorder.length - 1;        int p = i;        TreeNode node;        TreeNode root = new TreeNode(postorder[postorder.length - 1]);        Stack<TreeNode> stack = new Stack<>();        stack.push(root);        p--;        while (true) {            if (inorder[i] == stack.peek().val) { // inorder[i] is on top of stack, pop stack to get its parent to get to left side                if (--i < 0) break;                node = stack.pop();                if (!stack.isEmpty() && inorder[i] == stack.peek().val) {// continue pop stack to get to left side                    continue;                }                node.left = new TreeNode(postorder[p]);                stack.push(node.left);            } else { // inorder[i] is not on top of stack, postorder[p] must be right child                node = new TreeNode(postorder[p]);                stack.peek().right = node;                stack.push(node);            }            p--;        }        return root;    }