Question

Source

• lintcode: (245) Subtree

You have two every 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.

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
Note
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.

题解

判断 T2 是否是 T1 的子树，首先应该在 T1 中找到 T2 的根节点，找到根节点后两棵子树必须完全相同。所以整个思路分为两步走：找根节点，判断两棵树是否全等。咋看起来极其简单，但实际实现中还是比较精妙的，尤其是递归的先后顺序及条件与条件或的处理。

Java

/**
 * 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, T2: The roots of binary tree.
   * @return: True if T2 is a subtree of T1, or false.
   */
  public boolean isSubtree(TreeNode T1, TreeNode T2) {
    if (T2 == null) return true;
    if (T1 == null) return false;
    return identical(T1, T2) || isSubtree(T1.left, T2) || isSubtree(T1.right, T2);
  }

  private boolean identical(TreeNode T1, TreeNode T2) {
    if (T1 == null && T2 == null) return true;
    if (T1 == null || T2 == null) return false;
    if (T1.val != T2.val) return false;
    return identical(T1.left, T2.left) && identical(T1.right, T2.right);
  }
}

源码分析

这道题的异常处理相对 trick 一点，需要理解 null 对子树的含义。另外需要先调用 identical 再递归调用 isSubtree 判断左右子树的情况。方法 identical 中调用 .val 前需要判断是否为 null, 而后递归调用判断左右子树是否 identical。

复杂度分析

identical 的调用，时间复杂度近似 O(n)O(n)O(n), 查根节点的时间复杂度随机，平均为 O(m)O(m)O(m), 故总的时间复杂度可近似为 O(mn)O(mn)O(mn).

Reference

• LintCode: Subtree