【剑指offer】面试题26：树的子结构

完整代码地址

完整代码地址

题目

输入两棵二叉树A，B，判断B是不是A的子结构。
（ps：我们约定空树不是任意一个树的子结构）

思路

1. 找到树A中与树B根节点值相同的节点，设树A中该节点为subRoot
2. 在遍历树B的同时，遍历subRoot，如果遍历完树B发现subRoot中没有与B不一致的地方，则B是A的子结构
3. 遍历树B时，发现subRoot中有与树B结构不一致的地方，则退出遍历
4. 重复步骤1，找到树A中与树B根节点下一个值相同的节点

我觉得我这题的代码写的挺妙的哈哈哈

代码

public static class TreeNode {    public int val = 0;    public TreeNode left = null;    public TreeNode right = null;    public TreeNode(int val) {        this.val = val;    }}public boolean HasSubtree(TreeNode root1, TreeNode root2) {    if(root1 == null || root2 == null)        return false;    return findRootWithValue(root1, root2.val, root2);}/** * 遍历root1，当找到某个节点值与value相同时，遍历root2和子树root做比较 */public boolean findRootWithValue(TreeNode root, int value, TreeNode root2) {    if(root == null)        return false;    // 找到了根节点值一样的，并且root中包含root2    if(root.val == value && compare(root2, root)) {        return true;    }    return findRootWithValue(root.left, value, root2) ||             findRootWithValue(root.right, value, root2);}public boolean compare(TreeNode root, TreeNode rootToBeCompared) {    if(root == null)        return true;    if(rootToBeCompared == null)        return false;    if(root.val != rootToBeCompared.val)        return false;    return compare(root.left, rootToBeCompared.left) &&            compare(root.right, rootToBeCompared.right);}

测试

public static void main(String[] args) {    test1();    test2();    test3();    test4();}/** * 功能测试 *     1 *    / \ *   2   3 *  /\   /\ * 4  5 6  7 */private static void test1() {    TreeNode root = 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);    root.left = node2;    root.right = node3;    node2.left = node4;    node2.right = node5;    node3.left = node6;    node3.right = node7;    _26_SubStructureInTree ssit = new _26_SubStructureInTree();    boolean b = ssit.HasSubtree(root, node3);    MyTest.equal(b, true);    TreeNode root2 = new TreeNode(1);    root2.left = new TreeNode(4);    root2.right = new TreeNode(3);    b = ssit.HasSubtree(root, root2);    MyTest.equal(b, false);}/** * 边界测试  * 1.root2只有一个节点 */private static void test2() {    TreeNode root = 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);    root.left = node2;    root.right = node3;    node2.left = node4;    node2.right = node5;    node3.left = node6;    node3.right = node7;    _26_SubStructureInTree ssit = new _26_SubStructureInTree();    TreeNode root2 = new TreeNode(6);    boolean b = ssit.HasSubtree(root, root2);    MyTest.equal(b, true);    TreeNode root3 = new TreeNode(8);    b = ssit.HasSubtree(root, root3);    MyTest.equal(b, false);}/** * 极端测试  * 1.root1为 null * 2.root2为 null * 3.root1和root2都为null */private static void test3() {    TreeNode root1 = new TreeNode(1);    TreeNode root2 = new TreeNode(2);    _26_SubStructureInTree ssit = new _26_SubStructureInTree();    boolean b = ssit.HasSubtree(null, root2);    MyTest.equal(b, false);    b = ssit.HasSubtree(root1, null);    MyTest.equal(b, false);    b = ssit.HasSubtree(null, null);    MyTest.equal(b, false);}/** * 第一次没通过的case  */private static void test4() {    TreeNode node1 = new TreeNode(8);    TreeNode node2 = new TreeNode(8);    TreeNode node3 = new TreeNode(7);    TreeNode node4 = new TreeNode(9);    TreeNode node5 = new TreeNode(2);    TreeNode node6 = new TreeNode(4);    TreeNode node7 = new TreeNode(7);    node1.left = node2;    node1.right = node3;    node2.left = node4;    node2.right  =node5;    node5.left = node6;    node5.right = node7;    _26_SubStructureInTree ssit = new _26_SubStructureInTree();    boolean b = ssit.HasSubtree(node1, node2);    MyTest.equal(b, true);}