101. Symmetric Tree

题目：

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

For example, this binary tree is symmetric:

    1
   / 
  2   2
 /  / 
3  4 4  3

But the following is not:

    1
   / 
  2   2
      
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

Hide Tags

 Tree Depth-first Search

链接：http://leetcode.com/submissions/detail/11328058/

一刷

recursive

def isSymmetric(p, q):
    if not p and not q:
        return True
    if not p or not q:
        return False
    
    return p.val == q.val and isSymmetric(p.left, q.right) and isSymmetric(p.right, q.left)
        
class Solution(object):
    def isSymmetric(self, root):
        if not root:
            return True
        return isSymmetric(root.left, root.right)

iterative，利用tuple来成对比较，答案是抄的。。。

class Solution(object):
    def isSymmetric(self, root):
        if not root:
            return True
        stack = [(root.left, root.right)]
        
        while stack:
            l_elem, r_elem = stack.pop()
            if not l_elem and not r_elem:
                continue
            if not l_elem or not r_elem:
                return False
            if l_elem.val == r_elem.val:
                stack.extend([(l_elem.left, r_elem.right), (l_elem.right, r_elem.left)])
            else:
                return False
        return True

参考：

https://leetcode.com/discuss/14944/recursively-and-iteratively-solution-in-python

2/16/2017, Java

performance不够好，二刷再解决

public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null || root.left == null && root.right == null) return true;
        return isSymmetric(root.left, root.right);
    }
    private boolean isSymmetric(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;
        if (isSymmetric(p.left, q.right) && isSymmetric(p.right, q.left)) return true;
        return false;
    }
}