Difficulty : easy
Given two binary trees, write a function to check if they are equal or not.
Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
这道题一开始没想出来，因为tag是DFS, 我以为要写很复杂的recursion，至少是带helper method那种。但看了答案发现题目原本很简单，先比较root的val是不是相等，再比较左右子树isSameTree, 要都为真才能返回true.
/**
* 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.val != q.val){
return false;
}
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}
}
二刷，做到Symmetric Tree的时候回来做的，无压力ac
/**
* 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;
} else if (p == null || q == null){
return false;
}
if (p.val != q.val){
return false;
}
if (!isSameTree(p.left, q.left)){
return false;
}
if (!isSameTree(p.right, q.right)){
return false;
}
return true;
}
}
