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算法笔记: 力扣#100 相同的树

admin 11月 12, 2020 0

问题描述

解法

分析

Python 实现

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class :
    def isSameTree(self, p, q):
        """
        :type p: TreeNode
        :type q: TreeNode
        :rtype: bool
        """
        if p is None and q is None:
            return True
        if (p is None and q is not None) or (p is not None and q is None):
            return False
        if p.val != q.val:
            return False
        if self.isSameTree(p.left, q.left):
            return self.isSameTree(p.right, q.right)
        return False

Java 实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class  {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if(p==null && q==null){ return true; }
        if((p==null && q!=null) || (p!=null && q==null)){ return false; }
        if(p.val != q.val){ return false; }
        if(isSameTree(p.left, q.left)){ return isSameTree(p.right, q.right); }
        return false;
    }
}

时间复杂度

O(n).

空间复杂度

O(1).

链接

100. Same Tree
100. 相同的树
 (English version) Algorithm Notes: Leetcode#100 Same Tree