Given two binary trees, write a function to check if they are the same or not.node
Two binary trees are considered the same if they are structurally identical and the nodes have the same value.ide
Example 1:spa
Input: 1 1 / \ / \ 2 3 2 3 [1,2,3], [1,2,3] Output: true
Example 2:it
Input: 1 1 / \ 2 2 [1,2], [1,null,2] Output: false
Example 3:io
Input: 1 1 / \ / \ 2 1 1 2 [1,2,1], [1,1,2] Output: false
# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def isSameTree(self, p, q): """ :type p: TreeNode :type q: TreeNode :rtype: bool """ # DFS if p and q: return p.val == q.val and self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right) else: return p == q
迭代解法:function
public boolean isSameTree(TreeNode p, TreeNode q) { Stack<TreeNode> stack_p = new Stack <> (); Stack<TreeNode> stack_q = new Stack <> (); if (p != null) stack_p.push( p ) ; if (q != null) stack_q.push( q ) ; while (!stack_p.isEmpty() && !stack_q.isEmpty()) { TreeNode pn = stack_p.pop() ; TreeNode qn = stack_q.pop() ; if (pn.val != qn.val) return false ; if (pn.right != null) stack_p.push(pn.right) ; if (qn.right != null) stack_q.push(qn.right) ; if (stack_p.size() != stack_q.size()) return false ; if (pn.left != null) stack_p.push(pn.left) ; if (qn.left != null) stack_q.push(qn.left) ; if (stack_p.size() != stack_q.size()) return false ; } return stack_p.size() == stack_q.size() ; }
判斷size是否同!class