LeetCode题解-105-Construct Binary Tree from Preorder and Inorder Traversal

原题

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:You may assume that duplicates do not exist in the tree. 

原题链接：https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

解题思路

前序遍历与中序遍历的结构为：

前序遍历：[ 根节点， 左子树节点X个， 右子树节点Y个 ]；

中序遍历：[ 左子树节点X个， 根节点， 右子树节点Y个 ]；

因此，解题思路为：

1、前序遍历的第一个节点为根节点；

2、中序遍历中可找到根节点，在根节点之前的为左子树的节点，在根节点之后的为右子树的节点；

3、知道了左子树的节点个数与右子树的节点个数后，可以找到前序遍历中左子树节点与右子树节点；

4、递归终止条件为前序遍历仅有一个节点，此时该节点即为根节点。

代码1

该实现参考： https://discuss.leetcode.com/topic/3695/my-accepted-java-solution/2，比较简洁。

public class Solution105_recursive {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
       return buildTreeHelper(0, 0, inorder.length - 1, preorder, inorder);
    }

    private TreeNode buildTreeHelper(int preStart, int inStart, int inEnd, int[] preorder, int[] inorder) {
        if (preStart > preorder.length - 1 || inStart > inEnd)
            return  null;

        TreeNode root = new TreeNode(preorder[preStart]);
        int inIndex = 0;
        for (int i = inStart; i <= inEnd; i++){
            if (inorder[i] == root.val)
                inIndex = i;
        }

        root.left = buildTreeHelper(preStart + 1, inStart, inIndex - 1, preorder, inorder);
        root.right = buildTreeHelper(preStart + inIndex - inStart + 1, inIndex + 1, inEnd, preorder, inorder);
        return root;
    }
}

代码2

为自己最初的解法，

public class Solution105_recursive {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder.length == 0)
            return null;
        else if (preorder.length == 1)
            return new TreeNode(preorder[0]);

        int rootIndexOfInorder = getIndexFromArray(preorder[0], inorder);
        int[] leftInOrder = getSubArray(inorder, 0, rootIndexOfInorder - 1);
        int[] rightInOrder =  getSubArray(inorder, rootIndexOfInorder + 1, inorder.length - 1);
        int[] leftPreOrder = getSubArray(preorder, 1, leftInOrder.length);
        int[] rightPreOrder = getSubArray(preorder, leftInOrder.length + 1, leftInOrder.length + rightInOrder.length);

        TreeNode root = new TreeNode(preorder[0]);
        root.left = buildTree(leftPreOrder, leftInOrder);
        root.right = buildTree(rightPreOrder, rightInOrder);

        return root;
    }

    private int[] getSubArray(int[] array, int start, int end) {
        if (start > end)
            return new int[0];

        int[] subArray  = new int[end - start + 1];
        int indexOfSubArray = 0, indexOfArray = start;
        while (indexOfArray != end)
            subArray[indexOfSubArray++] = array[indexOfArray++];
        subArray[indexOfSubArray] = array[indexOfArray];

        return subArray;
    }

    private int getIndexFromArray(int val, int[] array) {
        int index = 0;
        while (val != array[index++]);

        return --index;
    }

}