Construct Binary Tree from Inorder and Postorder Traversal

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

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

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

  3
 / \
9  20
  /  \
 15   7

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {number[]} inorder
 * @param {number[]} postorder
 * @return {TreeNode}
 */
var buildTree = function(inorder, postorder) {
  return dfs(inorder.length - 1, 0, inorder.length - 1);

  function dfs(index, startPos, endPos) {
    if (startPos > endPos)
      return null;

    var node = new TreeNode(postorder[index]);
    var pos = inorder.indexOf(postorder[index], startPos);

    node.left = dfs(index - (endPos - pos) - 1, startPos, pos - 1);
    node.right = dfs(index - 1, pos + 1, endPos);

    return node;
  }
};

Construct Binary Tree from 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.

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

  3
 / \
9  20
  /  \
 15   7

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {number[]} preorder
 * @param {number[]} inorder
 * @return {TreeNode}
 */
var buildTree = function(preorder, inorder) {
  return dfs(0, 0, preorder.length - 1);

  // the subTree values whose node value is preorder[index]
  // inorder[startPos] - inorder[endPos]
  function dfs(index, startPos, endPos) {
    if (startPos > endPos)
      return null;

    var node = new TreeNode(preorder[index]);
    var pos = inorder.indexOf(preorder[index], startPos);

    // node's left subNode's value is preorder[index + 1]
    // node's right subNode's value is preorder[index + pos - startPos + 1]
    node.left = dfs(index + 1, startPos, pos - 1);
    node.right = dfs(index + pos - startPos + 1, pos + 1, endPos);

    return node;
  }
};

Construct the Rectangle

For a web developer, it is very important to know how to design a web page’s size. So, given a specific rectangular web page’s area, your job by now is to design a rectangular web page, whose length L and width W satisfy the following requirements:

1. The area of the rectangular web page you designed must equal to the given target area.

2. The width W should not be larger than the length L, which means L >= W.

3. The difference between length L and width W should be as small as possible.

You need to output the length L and the width W of the web page you designed in sequence.

Example:

Input: 4
Output: [2, 2]
Explanation: The target area is 4, and all the possible ways to construct it are [1,4], [2,2], [4,1]. 
But according to requirement 2, [1,4] is illegal; according to requirement 3,  [4,1] is not optimal compared to [2,2]. So the length L is 2, and the width W is 2.

Note:

1. The given area won’t exceed 10,000,000 and is a positive integer
2. The web page’s width and length you designed must be positive integers.