给出二叉树的中序和后序遍历，构建出二叉树

假设二叉树中序遍历为：

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

后序遍历为

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

那么，如何构建还原出这颗二叉树？类似问题还有：给出二叉树的先序和中序遍历如何还原出二叉树

算法思想：

• 后序遍历中的最后一个元素为二叉树的根节点，这里是数字1
• 在中序遍历集合中查找数字1，记作位置i，注意i的左边是左子树，右边是右子树
• 假设在步骤2中，有X个元素在i的左侧，那么从后序遍历集合中取前X个元素，得到的结果子集合即为左子树的后序遍历，类似的，i的右侧有Y个元素，那么，从后续遍历集合中取接下来的Y个元素，得到的结果即为右子树的后序遍历
• 通过步骤2和步骤3构建出根节点的左子树和右子树

算法实现：

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;
    }
}

输出结果

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/