LeetCode105 construct-binary-tree-from-preorder-and-inorder-traversal（从前序和中序重建二叉树）

题目

Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

 Example 1: 
Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

 Example 2: 
Input: preorder = [-1], inorder = [-1]
Output: [-1]

 Constraints: 
 1 <= preorder.length <= 3000 
 inorder.length == preorder.length 
 -3000 <= preorder[i], inorder[i] <= 3000 
 preorder and inorder consist of unique values. 
 Each value of inorder also appears in preorder. 
 preorder is guaranteed to be the preorder traversal of the tree. 
 inorder is guaranteed to be the inorder traversal of the tree. 

方法

递归法

注意左子树的长度size计算

• 时间复杂度：O()
• 空间复杂度：O()

class Solution {
    private Map<Integer, Integer> indexMap;
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        int n = preorder.length;
        indexMap = new HashMap<Integer, Integer>();
        for (int i = 0; i < n; i++) {
            indexMap.put(inorder[i], i);
        }
        return myBuildTree(preorder, inorder, 0, n - 1, 0, n - 1);
    }
    public TreeNode myBuildTree(int[] preorder, int[] inorder, int preorder_left, int preorder_right, int inorder_left, int inorder_right) {
        if (preorder_left > preorder_right) {
            return null;
        }
        int inorder_root = indexMap.get(preorder[preorder_left]);
        TreeNode root = new TreeNode(preorder[preorder_left]);
        int size_left_subtree = inorder_root - inorder_left;
        root.left = myBuildTree(preorder, inorder, preorder_left + 1, preorder_left + size_left_subtree, inorder_left, inorder_root - 1);
        root.right = myBuildTree(preorder, inorder, preorder_left + size_left_subtree + 1, preorder_right, inorder_root + 1, inorder_right);
        return root;
    }
}