leetcode——101. 对称二叉树

class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:return True
        def Tree(p,q):
            if not p and not q:return True
            if p and q and p.val==q.val:
                return Tree(p.left,q.right) and Tree(p.right,q.left)
            return False
        return Tree(root.left,root.right)

执行用时 :24 ms, 在所有 python 提交中击败了76.38%的用户

内存消耗 :12 MB, 在所有 python 提交中击败了14.26%的用户

 

 

——2019.11.10

 

还是运用递归的思想，挺简单好用的。

public boolean isSymmetric(TreeNode root) {
        if(root == null){
            return true;
        }
        return isSymmetric(root.left,root.right);
    }

    private boolean isSymmetric(TreeNode left, TreeNode right) {
        if(left == null && right == null){
            return true;
        }else if(left == null || right == null){
            return false;
        }else if(left.val != right.val){
            return false;
        }
        return isSymmetric(left.left,right.right) && isSymmetric(left.right,right.left);
    }

——2020.7.1