题目链接

递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder.length == 0 || inorder.length == 0){
            return null;
        }
        return buildHelper(preorder, 0, preorder.length, inorder, 0, inorder.length);
    }

    public TreeNode buildHelper(int[] preorder, int preorderStart, int preorderEnd, int[] inorder, int inorderStart, int inorderEnd){
        if(preorderStart == preorderEnd){
            return null;
        }
        int rootVal = preorder[preorderStart];
        TreeNode root = new TreeNode(rootVal);
        int middleIndex;
        for(middleIndex = inorderStart; middleIndex < inorderEnd; middleIndex++){
            // rootval
            if(inorder[middleIndex] == rootVal){
                break;
            }
        }
        // 对中序遍历处理
        int leftInorderStart = inorderStart; 
        int leftInorderEnd = middleIndex;
        int rightInorderStart = middleIndex + 1;
        int rightInorderEnd = inorderEnd;
        // 对前序遍历处理
        int leftPreorderStart = preorderStart + 1; 
        int leftPreorderEnd = preorderStart + 1 + middleIndex - inorderStart;
        int rightPreorderStart = preorderStart + 1 + middleIndex - inorderStart;
        int rightPreorderEnd = preorderEnd;

        root.left = buildHelper(preorder, leftPreorderStart, leftPreorderEnd , inorder, leftInorderStart, leftInorderEnd);
        root.right = buildHelper(preorder,rightPreorderStart, rightPreorderEnd, inorder, rightInorderStart, rightInorderEnd);

        return root;
    }
}