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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）
代码实现方式一：
/** public class TreeNode { int val = 0; TreeNode left = null; TreeNode right = null; public TreeNode(int val) { this.val = val; } } */ public class Solution { public boolean HasSubtree(TreeNode root1,TreeNode root2) { if (root1 == null || root2 == null) { return false; } return isSubTree(root1, root2) || HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2); } public boolean isSubTree(TreeNode root1, TreeNode root2) { if (root2 == null) return true; if (root1 == null) return false; if (root1.val == root2.val) { return isSubTree(root1.left, root2.left) && isSubTree(root1.right, root2.right); } else return false; } }
代码实现方式二：
/** public class TreeNode { int val = 0; TreeNode left = null; TreeNode right = null; public TreeNode(int val) { this.val = val; } } */ public class Solution { public boolean HasSubtree(TreeNode root1,TreeNode root2) { boolean result = false; if (root1 != null && root2 != null) { if (root1.val == root2.val) { result = isSubTree(root1, root2); } if (!result) { result = HasSubtree(root1.left, root2); } if (!result) { result = HasSubtree(root1.right, root2); } } return result; } public boolean isSubTree(TreeNode root1, TreeNode root2) { if (root1 == null && root2 != null) return false; if (root2 == null) { return true; } if (root1.val != root2.val) { return false; } return isSubTree(root1.left, root2.left) && isSubTree(root1.right, root2.right); } }
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原文地址：http://www.cnblogs.com/SaraMoring/p/5813877.html
