LeetCode 572. Subtree of Another Tree

原题链接在这里：https://leetcode.com/problems/subtree-of-another-tree/#/description

题目：

Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node's descendants. The tree s could also be considered as a subtree of itself.

Example 1:
Given tree s:

     3
    / 
   4   5
  / 
 1   2

Given tree t:

   4 
  / 
 1   2

Return true, because t has the same structure and node values with a subtree of s.

Example 2:
Given tree s:

     3
    / 
   4   5
  / 
 1   2
    /
   0

Given tree t:

   4
  / 
 1   2

Return false.

题解：

Traverse s 的每个节点，看把该节点当成root的subtree是否与t identical.

如果s 和 t本身都是null, 那么t也算s的subtree, 因为null 是本身null 的subtree.

Time Complexity: O(m*n). m是s的节点数. n是t的节点数. 对于s的每一个节点都做了一次以该节点为root的subtree 与 t的比较.

Space: O(logm). stack space, 一直找不到的时候会走到s的底部.

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        if(isSame(s, t)){
            return true;
        }
        return s!=null && (isSubtree(s.left, t) || isSubtree(s.right,t ));
    }
    
    private boolean isSame(TreeNode s, TreeNode t){
        if(s == null && t == null){
            return true;
        }
        
        if(s == null || t == null){
            return false;
        }
        
        if(s.val != t.val){
            return false;
        }
        
        return isSame(s.left, t.left) && isSame(s.right, t.right);
    }
}

对s 和 t 分别做preorder traversal, 然后看t得到的string 是不是 s得到string的 substring.

遇到null TreeNode时要添加特别符号, 不然就不能区分 123, null, null 和 1, 2, 3. 也无法区分 1, null, 2 和 1, 2, null. 这两种情况.

Time Complexity: O(m+n). m是s的节点数. n是t的节点数.

Space: O(Math.max(logm, logn)). Stack space.

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        String sPreorder = preorderTraversal(s);
        String tPreorder = preorderTraversal(t);
        
        return sPreorder.contains(tPreorder);
    }
    
    private String preorderTraversal(TreeNode root){
        StringBuilder sb = new StringBuilder();
        Stack<TreeNode> stk = new Stack<TreeNode>();
        stk.push(root);
        while(!stk.isEmpty()){
            TreeNode tn = stk.pop();
            if(tn == null){
                sb.append(",#");
            }else{
                sb.append(","+tn.val);
                stk.push(tn.left);
                stk.push(tn.right);
            }
        }
        return sb.toString();
    }
}

类似Same Tree, Binary Tree Preorder Traversal.