【剑指offer】面试题7：重建二叉树

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完整代码地址

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中都不含重复的数字。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}，则重建二叉树并返回。

思路

前序遍历数组为pre[]，中序遍历数组为in[]
1. pre[]中首个点为root，设值为value
2. 在in[]中寻找值为value的点
3. in[]，值为value左边的子数组为左子树，右边的子数组为右子树
4. 根据子数组的长度在pre[]中确定左子树和右子树
5. 根据步骤4可以得到两个子数组，再重复步骤1
可以使用递归解决

代码

public static class TreeNode {    public int val;    public TreeNode left;    public TreeNode right;    public TreeNode(int x) {        this.val = x;    }}private static final int LEFT = 0;private static final int RIGHT = 1;/** * 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中都不含重复的数字。 *  * @param pre   前序遍历 * @param in    中序遍历 * @return  该二叉树 */public static TreeNode reConstructBinaryTree(int[] pre, int[] in) {    if(pre.length == 0 || in.length == 0)        return null;    TreeNode root = new TreeNode(0);    buildTree(root, root, pre, 0, pre.length, in, 0, in.length, LEFT);    if(root == null)         return null;    root = root.left;    return root;}public static void buildTree(TreeNode root, TreeNode sonRoot, int[] pre, int preStart, int preEnd,        int[] in, int inStart, int inEnd, int leftOrRight) {    if(sonRoot == null || preStart >= preEnd || inStart >= inEnd)        return;    TreeNode newRoot = null;    int value = pre[preStart];  // 根节点的值     int leftLen = 0;            // 左子树节点个数    int i;    for(i = inStart; i < inEnd; ++i) {        if(in[i] == value) {            leftLen = i - inStart;            break;        }    }    // 前序遍历和中序遍历不匹配    if(i == inEnd) {        root = null;        return;    }    if(leftOrRight == LEFT) {        sonRoot.left = new TreeNode(value);        newRoot = sonRoot.left;    }    else {        sonRoot.right = new TreeNode(value);        newRoot = sonRoot.right;    }    buildTree(root, newRoot, pre, preStart+1, preStart+1+leftLen, in, inStart, inStart+leftLen, LEFT);    buildTree(root, newRoot, pre, preStart+1+leftLen, preEnd, in, inStart+leftLen+1, inEnd, RIGHT);}

测试

public static void main(String[] args) {    //testForPreOrderAndInOrder();    test1();    test2();    test3();    test4();    test5();    test6();    test7();}/** * 普通二叉树1：完全二叉树 *      1 *     / \ *    2   3 *   / \ / \ *  4  5 6  7  *  pre: 1,2,4,5,3,6,7  *  in : 4,2,5,1,6,3,7 */private static void test1() {    System.out.println("==============test1==============");    TreeNode root = _07_ConstructBinaryTree.reConstructBinaryTree(            new int[]{1,2,4,5,3,6,7}, new int[]{4,2,5,1,6,3,7});    preOrder(root);    System.out.println();    inOrder(root);    System.out.println();}/** * 普通二叉树2：不完全二叉树 *      1 *     / \ *    2   3 *   /     \ *  4       7  *  pre: 1,2,4,3,7 *  in : 4,2,3,7,1 */private static void test2() {    System.out.println("==============test2==============");    TreeNode root = _07_ConstructBinaryTree.reConstructBinaryTree(            new int[]{1,2,4,3,7}, new int[]{4,2,3,7,1});    preOrder(root);    System.out.println();    inOrder(root);    System.out.println();}/** * 特殊二叉树1：所有 节点都没有右子节点的二叉树 *       1 *      / *     2 *    / *   3 *  / * 4  * pre: 1,2,3,4 * in : 4,3,2,1 */private static void test3() {    System.out.println("==============test3==============");    TreeNode root = _07_ConstructBinaryTree.reConstructBinaryTree(            new int[]{1,2,3,4}, new int[]{4,3,2,1});    preOrder(root);    System.out.println();    inOrder(root);    System.out.println();}/** * 特殊二叉树2：所有 节点都没有左子节点的二叉树 *        1 *         \ *          2 *           \  *            3 *             \ *              4 * pre: 1,2,3,4 * in : 1,2,3,4 */private static void test4() {    System.out.println("==============test4==============");    TreeNode root = _07_ConstructBinaryTree.reConstructBinaryTree(            new int[]{1,2,3,4}, new int[]{1,2,3,4});    preOrder(root);    System.out.println();    inOrder(root);    System.out.println();}/** * 特殊二叉树3：只有一个节点的二叉树 *    1 * pre: 1 * in : 1 */private static void test5() {    System.out.println("==============test5==============");    TreeNode root = _07_ConstructBinaryTree.reConstructBinaryTree(            new int[]{1}, new int[]{1});    preOrder(root);    System.out.println();    inOrder(root);    System.out.println();}/** * 特殊输入测试1：数组长度为0 */private static void test6() {    System.out.println("==============test6==============");    TreeNode root = _07_ConstructBinaryTree.reConstructBinaryTree(            new int[0], new int[0]);    preOrder(root);    System.out.println();    inOrder(root);    System.out.println();}/** * 特殊输入测试2：前序和中序遍历不匹配 *      1 *     / \ *    2   3 *   /     \ *  4       7  *  pre: 1,2,4,3,7 *  in : 4,2,3,7,6 */private static void test7() {    System.out.println("==============test7==============");    TreeNode root = _07_ConstructBinaryTree.reConstructBinaryTree(            new int[]{1,2,4,3,7}, new int[]{4,2,3,7,6});    preOrder(root);    System.out.println();    inOrder(root);    System.out.println();}/** * 前序遍历 */private static void preOrder(TreeNode root) {    if(root == null)         return;    int value = root.val;    System.out.print(value + " ");    preOrder(root.left);    preOrder(root.right);}/** * 中序遍历 */private static void inOrder(TreeNode root) {    if(root == null)         return;    inOrder(root.left);    System.out.print(root.val + " ");    inOrder(root.right);}/** *         1 *       /   \ *      2     3 *     /     / \  *    4     5   6 *     \       / *      7     8 *   preOrder: 1,2,4,7,3,5,6,8 *   inOrder : 4,7,2,1,5,3,8,6 */private static void testForPreOrderAndInOrder() {    TreeNode node1 = new TreeNode(1);    TreeNode node2 = new TreeNode(2);    TreeNode node3 = new TreeNode(3);    TreeNode node4 = new TreeNode(4);    TreeNode node5 = new TreeNode(5);    TreeNode node6 = new TreeNode(6);    TreeNode node7 = new TreeNode(7);    TreeNode node8 = new TreeNode(8);    node1.left = node2;    node1.right = node3;    node2.left = node4;    node3.left = node5;    node3.right = node6;    node4.right = node7;    node6.left = node8;    preOrder(node1);    System.out.println();    inOrder(node1);    System.out.println();}

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