题目:
给定两个二叉树,编写一个函数来检验它们是否相同。
如果两个树在结构上相同,并且节点具有相同的值,则认为它们是相同的。
示例 1:
输入: 1 1 / \ / \ 2 3 2 3 [1,2,3], [1,2,3] 输出: true
示例 2:
输入: 1 1 / \ 2 2 [1,2], [1,null,2] 输出: false
示例 3:
输入: 1 1 / \ / \ 2 1 1 2 [1,2,1], [1,1,2] 输出: false
解题思路:
对两棵树同时进行前序遍历,判断对应的每个节点是否相等,注意判断null值。
代码实现:
递归版本:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public boolean isSameTree(TreeNode p, TreeNode q) { if (p == null && q == null) return true; if (p != null && q == null) return false; if (p == null && q != null) return false; if (p.val != q.val) return false; return isSameTree(p.left, q.left) && isSameTree(p.right, q.right); } }
非递归版本:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public boolean isSameTree(TreeNode p, TreeNode q) { Stack<TreeNode> pStack = new Stack<TreeNode>(); Stack<TreeNode> qStack = new Stack<TreeNode>(); if (!isEqual(p, q)) return false; else { while (p != null && q != null) { if (isEqual(p, q)) { if (p.right != null) { pStack.push(p.right); qStack.push(q.right); } p = p.left; q = q.left; if (p == null && q ==null && !pStack.isEmpty() && !qStack.isEmpty()) { p = pStack.pop(); q = qStack.pop(); } } else { return false; } } return isEqual(p, q); } } private boolean isEqual(TreeNode p, TreeNode q) { if (p == null && q == null) return true; else if (p == null && q != null) return false; else if (p != null && q == null) return false; else { if (p.val != q.val) return false; if (p.left == null && q.left != null) return false; else if (p.left != null && q.left == null) return false; if (p.right == null && q.right != null) return false; else if (p.right != null && q.right == null) return false; } return true; } }