245. Subtree【LintCode java】

Description

You have two very large binary trees: T1, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree of T1.

A tree T2 is a subtree of T1 if there exists a node n in T1 such that the subtree of n is identical to T2. That is, if you cut off the tree at node n, the two trees would be identical.

Example

T2 is a subtree of T1 in the following case:

       1                3
      / \              /
T1 = 2   3      T2 =  4
        /
       4

T2 isn‘t a subtree of T1 in the following case:

       1               3
      / \               T1 = 2   3       T2 =    4
        /
       4解题：判断一个二叉树是不是另一个二叉树的问题。思路还是很清晰的，但是有些细节还是要注意下的。两个递归，一个用来判断两个树是否完全一样，这个只要不一样就返回false。还有一个递归函数是用来返回最终结果的。区别在于，如果第二个函数判断不等，还有继续往T2的子树探索，知道找到一个与T1完全相等的，或者一直到最后也不存在这样的函数。比较容易错的是，在判断结点值的时候，要先判断结点是不是null（判空）。代码如下：

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param T1: The roots of binary tree T1.
     * @param T2: The roots of binary tree T2.
     * @return: True if T2 is a subtree of T1, or false.
     */
    public boolean equal(TreeNode T1, TreeNode T2){
        if(T1 == null && T2 == null){
            return true;
        }else if(T1 == null && T2 != null || T1 != null && T2 == null||T1.val != T2.val){
            return false;
        }else{
            return equal(T1.left, T2.left) && equal(T1.right, T2.right);
        }
    }
    public boolean isSubtree(TreeNode T1, TreeNode T2) {
        // write your code here
        if(T2 == null)
            return true;
        if(T1 == null)
            return false;
        if(equal(T1 , T2))
            return true;
        else return isSubtree(T1.left, T2)||isSubtree(T1.right, T2);
    }
}

原文地址：https://www.cnblogs.com/phdeblog/p/9300306.html