前序遍历和中序遍历树构造二叉树

public class Solution {
    /**
     * @param inorder: A list of integers that inorder traversal of a tree
     * @param postorder: A list of integers that postorder traversal of a tree
     * @return: Root of a tree
     */
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        // write your code here
        return childNode(preorder, 0, preorder.length, inorder, 0, inorder.length);
        
    }
    
    public TreeNode childNode(int[] preorder, int preStart, int preLen, int[] inorder, int inoStart, int inoLen){
        if(inoLen < 1){
            return null;
        }
        TreeNode node = new TreeNode(preorder[preStart]);
        int offet = 0;
        for(; offet < inoLen; offet++){
            if(node.val == inorder[inoStart + offet]){
                break;
            }
        }
        node.left = childNode(preorder, preStart + 1, offet, inorder, inoStart, offet);
        node.right = childNode(preorder, preStart + offet + 1, preLen - offet - 1, inorder,inoStart + offet + 1, inoLen - offet - 1);
        return node;
    }
}

解题思路：通过前序遍历可以找到根节点，然后遍历中序遍历数组找到根节点的位置，分别计算左右子树的节点， 通过递归不断计算子节点

文采不好，大家谅解