Title Description
Two binary inputs A and B, B is not determined substructure A's. In FIG right sub tree structure on the left
Ideas analysis
- A tree is traversed to find the B tree root value of the same node R & lt;
- A decision tree in the R-rooted subtree contains the same B-tree structure.
Test Case
- Functional test: Tree A and B are ordinary binary tree; B is a tree structure or a sub-tree A is not.
- Special input test: two binary tree - a pointer or two nullptr root node; all nodes of the binary tree or subtree are no left and right subtrees.
Java code
public class Offer26 {
public static void main(String[] args) {
test1();
test2();
test3();
}
public static boolean hasSubtree(TreeNode root1, TreeNode root2) {
return Solution1(root1, root2);
}
/**
* 在树A中找到与树B中根节点值一样节点
*
* @param root1
* @param root2
* @return
*/
private static boolean Solution1(TreeNode root1, TreeNode root2) {
boolean result = false;
if (root1 != null && root2 != null) {
if (isEqusl(root1.val, root2.val)) {
result = DoesTreeHaveTree2(root1, root2);
}
if (!result) {
result = Solution1(root1.left, root2);
}
if (!result) {
result = Solution1(root1.right, root2);
}
}
return result;
}
/**
* 此方法用于判断树A中**以R为根结点**的子树是否包含B树一样的结构。
*
* @param root1
* @param root2
* @return
*/
private static boolean DoesTreeHaveTree2(TreeNode root1, TreeNode root2) {
if (root2 == null) {
return true;
}
if (root1 == null) {
return false;
}
if (!isEqusl(root1.val, root2.val)) {
return false;
}
return DoesTreeHaveTree2(root1.left, root2.left) && DoesTreeHaveTree2(root1.right, root2.right);
}
/**
* 浮点数比较大小
*
* @param num1
* @param num2
* @return
*/
private static boolean isEqusl(float num1, float num2) {
final float THRESHOLD = 0.000001f;
if (Math.abs(num1 - num2) < THRESHOLD) {
return true;
} else {
return false;
}
}
private static void test1() {
}
private static void test2() {
}
private static void test3() {
}
}