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/*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
*
* 先从根节点开始判断,
* 1.如果根节点相同,则从头遍历
* 2.如果根不同,则从左右两个遍历,任意一个含有B树都可以,如果失败则递归
* doesTree1HasTree2函数负责递归两个遍历两个树,直到B树为空则返回true其余都是false
*
* */
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
public boolean HasSubTree(TreeNode root1, TreeNode root2){
boolean result = false;
if(root1 != null || root2 !=null){
if(root1.val == root2.val){
result = doesTree1HasTree2(root1, root2);
}
if (!result){ //如果上个节点失败,则继续比较tree1的左右子节点
result = doesTree1HasTree2(root1.left, root2) || doesTree1HasTree2(root1.right, root2);
}
}
return result;
}
public boolean doesTree1HasTree2(TreeNode root1, TreeNode root2){
if (root2 == null)
return true;
if (root1 == null)
return false;
if (root1.val != root2.val)
return false;
return doesTree1HasTree2(root1.left, root2.left) && doesTree1HasTree2(root1.right, root2.right);
}