Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1 / 2 2 / \ / 3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1 / 2 2 \ 3 3
recursively
class Solution { public boolean isSymmetric(TreeNode root) { return root==null||judge(root.left,root.right); } public boolean judge(TreeNode l,TreeNode r){ if(l==null||r==null) return l==r; if(l.val!=r.val) return false; return judge(l.left,r.right)&&judge(l.right,r.left); } }
iteratively
public boolean isSymmetric(TreeNode root) { if(root==null) return true; Stack<TreeNode> stack = new Stack<TreeNode>(); TreeNode left, right; if(root.left!=null){ if(root.right==null) return false; stack.push(root.left); stack.push(root.right); } else if(root.right!=null){ return false; } while(!stack.empty()){ if(stack.size()%2!=0) return false; right = stack.pop(); left = stack.pop(); if(right.val!=left.val) return false; if(left.left!=null){ if(right.right==null) return false; stack.push(left.left); stack.push(right.right); } else if(right.right!=null){ return false; } if(left.right!=null){ if(right.left==null) return false; stack.push(left.right); stack.push(right.left); } else if(right.left!=null){ return false; } } return true; }
Leetcode--101. Symmetric Tree(easy)
原文：https://www.cnblogs.com/albert67/p/10361101.html