給出二叉樹的中序和後序遍歷，構建出二叉樹

假設二叉樹中序遍歷爲：

int[] inOrder = { 4, 2, 5, 1, 6, 3, 7 };

後序遍歷爲

int[] postOrder = { 4, 5, 2, 6, 7, 3, 1 };

那麼，如何構建還原出這顆二叉樹？類似問題還有：給出二叉樹的先序和中序遍歷如何還原出二叉樹

算法思想：

• 後序遍歷中的最後一個元素爲二叉樹的根節點，這裏是數字1
• 在中序遍歷集合中查找數字1，記作位置i，注意i的左邊是左子樹，右邊是右子樹
• 假設在步驟2中，有X個元素在i的左側，那麼從後序遍歷集合中取前X個元素，得到的結果子集合即爲左子樹的後序遍歷，類似的，i的右側有Y個元素，那麼，從後續遍歷集合中取接下來的Y個元素，得到的結果即爲右子樹的後序遍歷
• 通過步驟2和步驟3構建出根節點的左子樹和右子樹

算法實現：

public class InorderPostOrderToTree {

    public static int pIndex = 0;

    public Node makeBTree(int[] inOrder, int[] postOrder, int iStart, int iEnd,
                          int postStart, int postEnd) {
        if (iStart > iEnd || postEnd > postEnd) {
            return null;
        }

        int rootValue = postOrder[postEnd];
        Node root = new Node(rootValue);
        pIndex++;

        if (iStart == iEnd) {
            return root;
        }
        int index = getInorderIndex(inOrder, iStart, iEnd, root.data);
        root.left = makeBTree(inOrder, postOrder, iStart, index - 1, postStart,
                postStart + index - (iStart + 1));
        root.right = makeBTree(inOrder, postOrder, index + 1, iEnd, postStart
                + index - iStart, postEnd - 1);
        return root;
    }

    public int getInorderIndex(int[] inOrder, int start, int end, int data) {
        for (int i = start; i <= end; i++) {
            if (inOrder[i] == data) {
                return i;
            }
        }
        return -1;
    }

    public void printInorder(Node root) {
        if (root != null) {
            printInorder(root.left);
            System.out.print(root.data + " ");
            printInorder(root.right);
        }
    }

    public static void main(String[] args) {
        int[] inOrder = {4, 2, 5, 1, 6, 3, 7};
        int[] postOrder = {4, 5, 2, 6, 7, 3, 1};
        InorderPostOrderToTree i = new InorderPostOrderToTree();
        Node x = i.makeBTree(inOrder, postOrder, 0, inOrder.length - 1, 0,
                postOrder.length - 1);
        System.out.println("inorder traversal of constructed tree:");
        i.printInorder(x);

    }
}

class Node {
    int data;
    Node left;
    Node right;

    public Node(int data) {
        this.data = data;
        left = null;
        right = null;
    }
}

輸出結果

inorder traversal of constructed tree:
4 2 5 1 6 3 7 

https://algorithms.tutorialhorizon.com/construct-a-binary-tree-from-given-inorder-and-postorder-traversal/