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
Note:
Bonus points if you could solve it both recursively and iteratively.
思路:朴素思想是基于层次遍历,然后判断每一层是否对称,然后再判断下一层。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public boolean isSymmetric(TreeNode root) { if(root == null) return true; Queue<TreeNode> layer = new LinkedList<TreeNode>(); ArrayList<TreeNode> compara = new ArrayList<TreeNode>(); layer.add(root); while(!layer.isEmpty()){ while(!layer.isEmpty()){ TreeNode nowNode = layer.poll(); if(nowNode == null) continue; compara.add(nowNode.left); compara.add(nowNode.right); } for(int left = 0,right = compara.size() - 1;left < right; left++,right--){ if(compara.get(left) != null && compara.get(right) != null &&compara.get(left).val == compara.get(right).val){ continue; }else if(compara.get(left) == null && compara.get(right) == null){ continue; }else{ System.out.println(); return false; } } layer.addAll(compara); compara.clear(); } return true; } }
当然,还有改进写法,可以减少一个辅助数组。
public boolean isSymmetric(TreeNode root) { Queue<TreeNode> q = new LinkedList<>(); q.add(root); q.add(root); while (!q.isEmpty()) { TreeNode t1 = q.poll(); TreeNode t2 = q.poll(); if (t1 == null && t2 == null) continue; if (t1 == null || t2 == null) return false; if (t1.val != t2.val) return false; q.add(t1.left); q.add(t2.right); q.add(t1.right); q.add(t2.left); } return true; }
public boolean isSymmetric(TreeNode root) { return isMirror(root, root); } public boolean isMirror(TreeNode t1, TreeNode t2) { if (t1 == null && t2 == null) return true; if (t1 == null || t2 == null) return false; return (t1.val == t2.val) && isMirror(t1.right, t2.left) && isMirror(t1.left, t2.right); }时空复杂度都是O(N),因为遍历所有节点,且使用了N的辅助空间(队列或者递归栈)。