105. Construct Binary Tree from Preorder and Inorder Traversal（由先序遍历和中序遍历构建二叉树）

题目描述

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

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

方法思路

Approach1:
由中序遍历和先序遍历确定二叉树
0.确定递归返回的临界条件。
if(start > end) return null;
1.根据先序遍历确定根结点：TreeNode root = new TreeNode(preorder[preIndex++]);因为先序遍历的特 性，可将preIndex设置为全局变量，很方便，而后序遍历与中序遍历的组合就无法使用全局变量了 。
2.在中序遍历的数组中确定根结点的索引：int index = map.get(root.val);
3.确定左子树与右子树的范围。
root.left = btHelper(preorder,map,start,index-1);
root.right = btHelper(preorder,map,index+1,end);
4.return root;

class Solution {
    //Runtime: 2 ms, faster than 97.75% 
    //Memory Usage: 37.2 MB, less than 86.50%
    int preIndex = 0;
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        
        if(preorder == null || preorder.length == 0) return null;
        Map<Integer,Integer> map = new HashMap();
        for(int i = 0; i < inorder.length; i++){
            map.put(inorder[i],i);
        }
        
        return btHelper(preorder,map,0, inorder.length-1);
        
        
    }
    
    private TreeNode btHelper(int[] preorder, Map<Integer,Integer> map, int start, int end){
        
        if(start > end) return null;
        //由先序遍历数组可以很容易的得到根结点的值
        TreeNode root = new TreeNode(preorder[preIndex++]);
        
        //index 是当前子树的根结点的索引，由此可以得到左右子树的索引范围，从而进行递归
        int index = map.get(root.val);
        
        root.left = btHelper(preorder,map,start,index-1);
        root.right = btHelper(preorder,map,index+1,end);
        
        return root;
        
    }
}

Approach2:
更加一般性的方法

class Solution {
    //Runtime: 7 ms, faster than 53.85%
    //Memory Usage: 37.2 MB, less than 82.21%
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return build(inorder, preorder, 0, 0, inorder.length-1);
    }
    
    public TreeNode build(int[] inorder, int[] preorder, int pre_start, int in_start, int in_end) {
        if (pre_start > preorder.length-1 || in_start > in_end) return null;
        
        TreeNode root = new TreeNode(preorder[pre_start]);
				
        int index = 0;
        for (int i = in_start; i <= in_end; i++) 
	        if (inorder[i] == root.val) 
                index = i;

        root.left = 
            build(inorder, preorder, pre_start + 1, in_start, index-1);
        root.right = 
            build(inorder, preorder, pre_start + (index - in_start + 1), index + 1, in_end);
     
        return root;
    }
}