LeetCode brushing notes _105. Construct a binary tree from the preorder and middle order traversal sequence

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105. Construct a binary tree from preorder and middle order traversal sequence

Other solutions or source code can be accessed: tongji4m3

description

The binary tree is constructed according to the pre-order and middle-order traversal of a tree.

Note:
You can assume that there are no duplicate elements in the tree.

例如，给出

前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树：

    3
   / \
  9  20
    /  \
   15   7

Ideas

For the array [lo,hi] traversed in pre-order and middle-order:

• The root node of the preorder is lo, and the root node of the middle order is i
• The left subtree of the preorder is [lo+1,i], the right subtree is [i+1,hi]
• The left subtree of the middle order is [lo,i-1], and the right subtree is [i+1,hi]

Therefore, the root node can be constructed first, and then his left and right nodes can be constructed according to the left and right arrays in the pre-order, middle-order, and continue to recurse...

detail

1. Note that the pre-order after several recursion, the middle-order array may not be symmetrical up and down, so their index index must be divided into preIndex, inIndex
2. Use map storage to speed up

Code

/**
 * look 用map存储加快速度
树中没有重复元素
用map存储inorder数组的值和对应索引
 方便找到根节点位置
 */
private Map<Integer, Integer> map = new HashMap<>();

public TreeNode buildTree(int[] preorder, int[] inorder)
{
    
    
    for (int i = 0; i < inorder.length; i++)
    {
    
    
        map.put(inorder[i], i);
    }
    return buildTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
}

private TreeNode buildTree(int[] preorder, int preLo, int preHi, int[] inorder, int inLo, int inHi)
{
    
    
    if (preLo > preHi || inLo > inHi)
    {
    
    
        return null;
    }
    //找到根节点在中序遍历的索引位置
    int inIndex = map.get(preorder[preLo]);
    //look,i在经过几次之后,只是对应了inorder的i,但是对于preorder,应该只是一个偏移量
    //inIndex - inLo代表左子树的个数,preIndex则是前序遍历左子树的最后一个元素
    int preIndex = preLo + inIndex - inLo;
    TreeNode root = new TreeNode(preorder[preLo]);
    root.left = buildTree(preorder, preLo + 1, preIndex, inorder, inLo, inIndex - 1);
    root.right = buildTree(preorder, preIndex + 1, preHi, inorder, inIndex + 1, inHi);
    return root;
}