Construct Binary Tree from Preorder and Inorder Traversal && Inorder and Postorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

可以利用preorder的第一个肯定是root的性质，同时找到此点在inorder中对应的位置，从而把这两个数组对应分割成左子树和右子树。

update: 在搜索Positions时可以用二分法，这样最坏Time: O(nlogn)

public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder == null || preorder.length == 0) {
            return null;
        }
        return build(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
    }
    
    private TreeNode build(int[] preorder, int prestart, int preend, 
                           int[] inorder, int instart, int inend) {
        if (preend < prestart || inend < instart) {
            return null;
        }               
        
        TreeNode root = new TreeNode(preorder[prestart]);
        int position = 0;
        for (int i = instart; i <= inend; i++) {
            if (inorder[i] == preorder[prestart]) {
                position = i;
            }
        }
        
        root.left = build(preorder, prestart + 1, prestart + position - instart,
                          inorder, instart, position - 1);
        root.right = build(preorder, prestart + position - instart + 1, preend,
                           inorder, position + 1, inend);
        return root;
    }
}

Given inorder and postorder traversal of a tree, construct the binary tree.

public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (inorder == null || inorder.length == 0) {
            return null;
        }
        
        return build(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);
    }
    
    private TreeNode build(int[] inorder, int instart, int inend,
                           int[] postorder, int poststart, int postend) {
        if (inend < instart || postend < poststart) {
            return null;
        }               
        
        TreeNode root = new TreeNode(postorder[postend]);
        int position = 0;
        for (int i = instart; i <= inend; i++) {
            if (inorder[i] == postorder[postend]) {
                position = i;
            }
        }
        
        root.left = build(inorder, instart, position - 1, 
                          postorder, poststart, poststart + position - instart - 1);
        root.right = build(inorder, position + 1, inend, 
                           postorder, poststart + position - instart, postend - 1);
        return root;
    }
}