105. Construct Binary Tree from Preorder and Inorder Traversal[Medium]

105. Construct Binary Tree from Preorder and Inorder Traversal

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.

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.

Example

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

思路

• 先序遍历（中左右）
• 中序遍历（左中右）
  所以先序数组的第一个值肯定是根节点，那在中序数组中该值位置前的所有值，都应该属于该节点的左子树，后面的都属于右子树
  这样就能得到一个该节点的左子树中序数组，一个该节点的右子树中序数组。
  根据左子树中序数组的长度，就能确定左子树先序数组的长度，然后偏移k个位置赋值，右子树先序数组同理

题解

        TreeNode root = new TreeNode();
        // 找到当前节点值在中序数组中的位置
        int index = 0;
        for (int cur : inorder) {
            index++;
            if (preorder[0] == cur)
                break;
        }
        // 没找到就返回
        if(index == 0)
            return null;
        int[] leftPre = new int[index - 1];
        int[] leftIn = new int[index - 1];
        int[] rightPre = new int[preorder.length - index];
        int[] rightIn = new int[preorder.length - index];
        for (int i = 0; i < leftPre.length; i++) {
            leftPre[i] = preorder[i + 1];
            leftIn[i] = inorder[i];
        }
        for (int i = 0; i < rightPre.length; i++) {
            rightPre[i] = preorder[index + i];
            rightIn[i] = inorder[index + i];
        }
        root.val = preorder[0];
        root.left = buildTree(leftPre, leftIn);
        root.right = buildTree(rightPre, rightIn);

        return root;