1. 程式人生 > >LeetCode:572. Subtree of Another Tree(判斷一棵樹是不是另外一棵樹的子樹)

LeetCode:572. Subtree of Another Tree(判斷一棵樹是不是另外一棵樹的子樹)

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

.


方法1:(還是利用遞迴的方式)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        if (s == null) {
            return false;
        }
        
        if (s.val == t.val && isSame(s, t)) {
            return true;
        }
        return isSubtree(s.left, t) || isSubtree(s.right, t);
    }
    
    private boolean isSame(TreeNode t1, TreeNode t2) {
        if (t1 == null && t2 == null) {
            return true;
        } else if (t1 == null || t2 == null) {
            return false;
        }
        if (t1.val == t2.val) {
            return isSame(t1.left, t2.left) && isSame(t1.right, t2.right);
        } else {
            return false;
        }
    }
}

時間複雜度:O(n^2)

空間複雜度:O(n)