101.Symmetric-Tree

101. Symmetric Tree

题目地址

https://leetcode.com/problems/swap-nodes-in-pairs/

题目描述

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.

代码

Approach #1 Recursive

Time complexity & Space complexity: O(n)

/**
 * 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) {
        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);
  }
}

Approach #2 Iterative

class Solution {
  public boolean isSymetric(TreeNode root) {
    Queue<TreeNode> queue = new LinkedList();
    queue.add(root);
    queue.add(root);

    while (!queue.isEmpty()) {
      TreeNode t1 = queue.poll();
      TreeNode t2 = queue.poll();
      if (t1 == null && t2 == null)    continue;
      if (t1 == null || t2 == null)    return false;
      if (t1.val != t2.val)    return false;
      queue.add(t1.left);
      queue.add(t2.right);
      queue.add(t1.right);
      queue.add(t2.left);
    }

    return true;
  }
}