LeetCode:Same Tree

Same Tree

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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;    }