剑指offer 面试题6 重建二叉树2

package algorithm.foroffer;import java.util.Arrays;/** * Created by liyazhou on 2017/5/23. * 面试题6 重建二叉树 * 题目： * 输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。 * 假设输入的前序遍历和中序遍历的结果都不含重复的数字。例如， * 输入前序遍历序列 {1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}， * 则重建出二叉树并输出它的头结点。 */class BinaryTreeNode{    private int value;  // id of BinaryTreeNode, unique    private BinaryTreeNode leftChildNode;    private BinaryTreeNode rightChildNode;    public BinaryTreeNode(){}    public BinaryTreeNode(int value){ this.value = value; }    public BinaryTreeNode(int value, BinaryTreeNode leftChildNode, BinaryTreeNode rightChildNode){        this.value = value;        this.leftChildNode = leftChildNode;        this.rightChildNode = rightChildNode;    }    public int getValue(){ return this.value; }    public BinaryTreeNode getLeftChildNode(){ return this.leftChildNode; }    public BinaryTreeNode getRightChildNode(){ return this.rightChildNode; }    public void setValue(int value){ this.value = value; }    public void setLeftChildNode(BinaryTreeNode leftChildNode){ this.leftChildNode = leftChildNode; }    public void setRightChildNode(BinaryTreeNode rightChildNode){ this.rightChildNode = rightChildNode; }}public class Test06 {    public BinaryTreeNode constructBinTree2(int[] preorder, int[] inorder){        // 判断输入数据        if(preorder == null || preorder.length == 0 ||           inorder == null || inorder.length == 0)            return null;        return constructBinTree(preorder, 0, preorder.length, inorder, 0, preorder.length);    }    // todo, 确定边界是关键点    private BinaryTreeNode constructBinTree(int[] preorder, int ps, int pe,                             int[] inorder, int is, int ie) {        // 递归终止条件，当开始位置大于结束位置时，则没有要处理的数据        if (ps >= pe || is >= ie) return null;        // 从前序遍历序列中取出根结点        int rootId = preorder[ps];        BinaryTreeNode root = new BinaryTreeNode(rootId);        // 根结点在中序遍历序列中的位置        int idxOfRoot = -1;        for (int i = is; i < ie; i++){            if(inorder[i] == rootId){                idxOfRoot = i;                break;            }        }        if (idxOfRoot == -1) {            System.out.println(String.format("ps, pe = %d, %d, preorder = %s", ps, pe, Arrays.toString(preorder)));            System.out.println(String.format("is, ie = %d, %d, inorder = %s", is, ie, Arrays.toString(inorder)));            throw new RuntimeException("先序遍历序列和中序遍历序列不匹配.");        }        // 递归构建当前根结点的左子树，左子树的结点个数是 idxOfRoot-is        // 先序遍历序列中，当前根结点的左子树的范围是 [ps+1, ps+(idxOfRoot-is)+1)，根据起始位置和偏移量计算范围        // 中序遍历序列中，当前根结点的左子树的范围是 [is, idxOfRoot)        BinaryTreeNode leftChildNode = constructBinTree(preorder, ps+1, ps+idxOfRoot-is+1, inorder, is, idxOfRoot);        root.setLeftChildNode(leftChildNode);        // 递归构建当前根结点的右子树，右子树的结点个数是 ie-idxOfRoot-1        // 先序遍历序列中，当前结点的右子树的范围是 [pe-(ie-idxOfRoot-1), pe)，根据终止位置和偏移量计算范围        // 中序遍历序列中，当前结点的右子树的范围是 [idxOfRoot+1, ie)        BinaryTreeNode rightChildNode = constructBinTree(preorder, pe-ie+idxOfRoot+1 , pe, inorder, idxOfRoot+1, ie);        root.setRightChildNode(rightChildNode);        return root;    }    public void previsit(BinaryTreeNode root){        if (root == null) return;        System.out.print(String.format("%s \t", root.getValue()));        previsit(root.getLeftChildNode());        previsit(root.getRightChildNode());    }    public void invisit(BinaryTreeNode root){        if (root == null) return;        invisit(root.getLeftChildNode());        System.out.print(String.format("%s \t", root.getValue()));        invisit(root.getRightChildNode());    }    public void postvisit(BinaryTreeNode root){        if (root == null) return;        postvisit(root.getLeftChildNode());        postvisit(root.getRightChildNode());        System.out.print(String.format("%s \t", root.getValue()));    }    public void visit(BinaryTreeNode root){        System.out.print("先序遍历结果：");        previsit(root);        System.out.println();        System.out.print("中序遍历结果：");        invisit(root);        System.out.println();        System.out.print("后序遍历结果：");        postvisit(root);        System.out.println("\n---------------------");    }    /**     * 普通二叉树     */    public void test1(){        int[] preorder = {1,2,4,7,3,5,6,8};        int[] inorder = {4,7,2,1,5,3,8,6};        BinaryTreeNode root = constructBinTree2(preorder, inorder);        visit(root);    }    /**     * 完全二叉树     */    public void test5(){        int[] preorder = {1, 2, 4, 5, 3, 6, 7};        int[] inorder = {4, 2, 5, 1, 6, 3, 7};        BinaryTreeNode root = constructBinTree2(preorder, inorder);        visit(root);    }    /**     * 所有结点都没有右孩子     */    public void test2(){        int[] preorder = {1, 2, 3, 4, 5};        int[] inorder = {5, 4, 3, 2, 1};        BinaryTreeNode root = constructBinTree2(preorder, inorder);        visit(root);    }    /**     * 所有结点都没有左孩子     */    public void test3(){        int[] preorder = {1, 2, 3, 4, 5};        int[] inorder = {1, 2, 3, 4, 5};        BinaryTreeNode root = constructBinTree2(preorder, inorder);        visit(root);    }    /**     * 只有一个结点     */    public void test4(){        int[] preorder = {1};        int[] inorder = {1};        BinaryTreeNode root = constructBinTree2(preorder, inorder);        visit(root);    }    /**     * 空指针     */    public void test6(){        BinaryTreeNode root = constructBinTree2(null, null);        visit(root);    }    /**     * 前序遍历和中序遍历不匹配     */    public void test7(){        int[] preorder = {1,3,2,4,7,5,6,8};        int[] inorder = {4,7,2,1,5,3,8,6};        BinaryTreeNode root = constructBinTree2(preorder, inorder);        visit(root);    }    public static void main(String[] args){        Test06 test = new Test06();        test.test1();        test.test2();        test.test3();        test.test4();        test.test5();        test.test6();        test.test7();    }}