[LeetCode] 101. 对称二叉树

题目链接 : https://leetcode-cn.com/problems/symmetric-tree/

题目描述:

给定一个二叉树，检查它是否是镜像对称的。

例如，二叉树 [1,2,2,3,4,4,3] 是对称的。

   1
   / \
  2   2
 / \ / \
3  4 4  3

但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的:

   1
   / \
  2   2
   \   \
   3    3

思路:

这道题和 上一题 100. 相同的树是一样的

我们只要比较root左右树是否对称(和是否相同的)就行了

思路一:递归

思路二:迭代

代码:

思路一:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSymmetric(self, root: TreeNode) -> 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)

java

/**
 * 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;
        return Tree(root.left, root.right);

    }

    public boolean Tree(TreeNode p, TreeNode q) {
        if (p == null && q == null) return true;
        if (p != null && q != null && p.val == q.val) return Tree(p.left, q.right) && Tree(p.right, q.left);
        else return false;
    }
}

思路二:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        if not root: return True
        def Tree(p, q):
            stack = [(q, p)]
            while stack:
                a, b = stack.pop()
                if not a and not b:
                    continue
                if a and b and a.val == b.val:
                    stack.append((a.left, b.right))
                    stack.append((a.right,b.left))
                else:
                    return False
            return True
        return Tree(root.left, root.right)

java

/**
 * 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;
        return Tree(root.left, root.right);

    }
     public boolean Tree(TreeNode p, TreeNode q) {
        Deque<TreeNode> stack1 = new LinkedList<>();
        Deque<TreeNode> stack2 = new LinkedList<>();
        stack1.push(p);
        stack2.push(q);
        while (!stack1.isEmpty() && !stack2.isEmpty()) {
            TreeNode a = stack1.pop();
            TreeNode b = stack2.pop();
            if (a == null && b == null) continue;
            if (a != null && b != null && a.val == b.val) {
                stack1.push(a.left);
                stack1.push(a.right);
                stack2.push(b.right);
                stack2.push(b.left);
            } else return false;
        }
        return stack1.isEmpty() && stack2.isEmpty();
    }
}