LeetCode101 Synmetrictree Java

Topic:

Given a binary tree, check whether it is a mirror of the itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1   /   2   2/\/3  4 4  3

But the following are not:

    1   /   2   2   \      3    3

Solving:

Symmetry is required that each layer from left to right is exactly equal to right to left, there are two solutions to a non-recursive, a recursive.

Non-recursive:

public static Boolean issymmetric (TreeNode root) {if (root==null) return true; Queue<treenode> left=new linkedlist<> (); Queue<treenode> right=new linkedlist<> (); if (root.left==null) Left.offer (root.left); if (root.right== NULL) Right.offer (Root.right), while (!left.isempty () &&!right.isempty ()) {TreeNode l=left.poll (); TreeNode R=right.poll (); if (l==null&&r==null) continue;if (l==null| | R==null) return false;if (L.val!=r.val) return False;left.offer (L.left); Left.offer (l.right); Right.offer (r.right); Right.offer (r.left);//reverse Insert form mirror}if (Left.isempty () &&right.isempty ()) return True;else {return false;}            

Recursion:

public Boolean isSymmetric2 (TreeNode root) {if (root==null) return true; else {return  issame (Root.left, root.right);}   }  public boolean issame (TreeNode Left,treenode right) {if (left==null&& right==null) return true; if (left==null| | Right==null) return false; if (Left.val!=right.val) return false; else {return Issame (Left.left, Right.right) &&issame (Left.right, right.left);}  }

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LeetCode101 Synmetrictree Java