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Given binary trees, write a function to check if they is equal or not.

The binary trees is considered equal if they is structurally identical and the nodes has the same value.

Test instructions: Judge whether the two trees are the same.

Ideas: One is recursion, one is to use the queue to do hierarchical traversal judgment.

/** * Definition for Binary tree * public class TreeNode {*     int val, *     TreeNode left, *     TreeNode right; *
   
    treenode (int x) {val = x;} *} */public class Solution {public    boolean issametree (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;    else return Issametree (P.left, Q.left) && issametree (P.right, q.right);}}}    
   

/** * Definition for Binary tree * public class TreeNode {* int val, * TreeNode left, * TreeNode right; TreeNode (int x) {val = x;} *} */public class Solution {Private Boolean Issamenode (TreeNode p, TreeNode q) {if (P = = NULL && q = = null) return True;if ((P! = NULL && q = = null) | | (p = = NULL && Q! = null) | | (P.val! = q.val))    return False;return true;}    public boolean issametree (TreeNode p, TreeNode q) {if (!issamenode (p, Q)) return false;        if (p = = NULL && q = = null) return true;    Linkedlist<treenode> left = new linkedlist<treenode> ();    linkedlist<treenode> right = new linkedlist<treenode> ();    Left.add (P);    Right.add (q);    while (Left.size ()! = 0 && right.size ()! = 0) {TreeNode ltmp = Left.peekfirst ();    TreeNode rtmp = Right.peekfirst ();    Left.removefirst ();        Right.removefirst ();    if (!issamenode (Ltmp.left, Rtmp.left)) return false; if (ltmp.left! = null &Amp;& Rtmp.left! = null) {Left.add (ltmp.left);    Right.add (Rtmp.left);    } if (!issamenode (Ltmp.right, Rtmp.right)) return false;    if (ltmp.right! = NULL && rtmp.right! = null) {Left.add (ltmp.right);    Right.add (Rtmp.right);    }} return true; }}

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