105. Construct Binary Tree from Preorder and Inorder Traversal

Question:

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

Thinking:

• Method：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder == null || inorder == null || preorder.length == 0 || inorder.length == 0) return null;
        Map<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i < inorder.length; i++)
            map.put(inorder[i], i);
        return createTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1, map);
    }
    public static TreeNode createTree(int[] preorder, int pStart, int pEnd, int[] inorder, int iStart, int iEnd, Map<Integer, Integer> map){
        if(pStart > pEnd || iStart > iEnd) return null;
        int rootVal = preorder[pStart];
        TreeNode root = new TreeNode(rootVal);
        int rootIndex = map.get(rootVal);
        int len = rootIndex - iStart;
        root.left = createTree(preorder, pStart + 1, pStart + len, inorder, iStart, rootIndex - 1, map);
        root.right = createTree(preorder, pStart + len + 1, pEnd, inorder, rootIndex + 1, iEnd, map);
        return root;
    }
}

二刷

1. 二刷的时候大概记得方法，但是还是参考了一刷时候的代码。一刷的时候其实也不是自己想出来的，而且还重复写了好几次。
2. 根据前序遍历的结果定位出在后序遍历中头结点的位置，通过计算计算可以得到左树和右树的长度，通过递归操作可以不断创建出左右结点。

   /**
    * Definition for a binary tree node.
    * public class TreeNode {
    *     int val;
    *     TreeNode left;
    *     TreeNode right;
    *     TreeNode(int x) { val = x; }
    * }
    */
   class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder == null || inorder == null) return null;
        Map<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i < inorder.length; i++)
            map.put(inorder[i], i);
        return createTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1, map);
    }
    private TreeNode createTree(int[] preorder, int pStart, int pEnd, int[] inorder, int iStart, int iEnd, Map<Integer, Integer> map){
        if(pStart > pEnd || iStart > iEnd) return null;
        TreeNode node = new TreeNode(preorder[pStart]);
        int index = map.get(preorder[pStart]);
        int len = index - iStart;
        node.left = createTree(preorder, pStart + 1, pStart + len, inorder, iStart, index - 1, map);
        node.right = createTree(preorder, pStart + len + 1, pEnd, inorder, index + 1, iEnd, map);
        return node;
    }
   }