LeetCode:Same Tree

Same Tree

 Total Accepted: 65115 Total
Submissions: 155756My Submissions

Question
 Solution 

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.

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/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/

方法一：递归

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;
}
boolean l=isSameTree(p.left,q.left);
boolean r=isSameTree(p.right,q.right);
return l&&r;
}

方法二：BFS（广度优先遍历）[LinkedList实现、add()、poll()]

  

public boolean isSameTree(TreeNode p, TreeNode q) {
if(p==null && q==null){
return true;
}
if(p==null || q==null){
return false;
}
LinkedList<TreeNode> queue1 = new LinkedList<TreeNode>();
LinkedList<TreeNode> queue2 = new LinkedList<TreeNode>();
queue1.add(p);
queue2.add(q);
while(!queue1.isEmpty()){
TreeNode n1 = queue1.poll();
TreeNode n2 = queue2.poll();
if(n1.val!=n2.val){
return false;
}
if(n1.left!=null && n2.left!=null){
queue1.add(n1.left);
queue2.add(n2.left);
}else if(!(n1.left==null && n2.left==null)){
return false;
}
if(n1.right!=null&&n2.right!=null){
queue1.add(n1.right);
queue2.add(n2.right);
}else if(!(n1.right==null && n2.right==null)){
return false;
}
}
return true;
}