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.
Solution1：遞歸對(duì)比
Time Complexity: O(N) Space Complexity: O(N) 遞歸緩存
Solution2：前序遍歷對(duì)比
思路： 除了前序的元素值瞻赶，前序的結(jié)構(gòu)也需要check静尼，是否同為Null或同存在辈双，通過(guò)push后(非null) check stack_p.size() != stack_q.size()
Time Complexity: O(N) Space Complexity: O(N)
Solution1 Code：
class Solution1 {
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 isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
return false;
}
}
Solution2 Code：
class Solution2 {
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() ;
}
}
