Question

101. 对称二叉树

English Version

题目描述

给你一个二叉树的根节点 root ， 检查它是否轴对称。

 

示例 1：

输入：root = [1,2,2,3,4,4,3]
输出：true

示例 2：

输入：root = [1,2,2,null,3,null,3]
输出：false

 

提示：

• 树中节点数目在范围 [1, 1000] 内
• -100 <= Node.val <= 100

 

进阶：你可以运用递归和迭代两种方法解决这个问题吗？

解法

方法一：递归

我们设计一个函数 $dfs(root1, root2)$，用于判断两个二叉树是否对称。答案即为 $dfs(root, root)$。

函数 $dfs(root1, root2)$ 的逻辑如下：

• 如果 $root1$ 和 $root2$ 都为空，则两个二叉树对称，返回 true；
• 如果 $root1$ 和 $root2$ 中只有一个为空，或者 $root1.val \neq root2.val$，则两个二叉树不对称，返回 false；
• 否则，判断 $root1$ 的左子树和 $root2$ 的右子树是否对称，以及 $root1$ 的右子树和 $root2$ 的左子树是否对称，这里使用了递归。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def isSymmetric(self, root: Optional[TreeNode]) -> bool:
    def dfs(root1, root2):
      if root1 is None and root2 is None:
        return True
      if root1 is None or root2 is None or root1.val != root2.val:
        return False
      return dfs(root1.left, root2.right) and dfs(root1.right, root2.left)

    return dfs(root, root)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  public boolean isSymmetric(TreeNode root) {
    return dfs(root, root);
  }

  private boolean dfs(TreeNode root1, TreeNode root2) {
    if (root1 == null && root2 == null) {
      return true;
    }
    if (root1 == null || root2 == null || root1.val != root2.val) {
      return false;
    }
    return dfs(root1.left, root2.right) && dfs(root1.right, root2.left);
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  bool isSymmetric(TreeNode* root) {
    function<bool(TreeNode*, TreeNode*)> dfs = [&](TreeNode* root1, TreeNode* root2) -> bool {
      if (!root1 && !root2) return true;
      if (!root1 || !root2 || root1->val != root2->val) return false;
      return dfs(root1->left, root2->right) && dfs(root1->right, root2->left);
    };
    return dfs(root, root);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func isSymmetric(root *TreeNode) bool {
  var dfs func(*TreeNode, *TreeNode) bool
  dfs = func(root1, root2 *TreeNode) bool {
    if root1 == nil && root2 == nil {
      return true
    }
    if root1 == nil || root2 == nil || root1.Val != root2.Val {
      return false
    }
    return dfs(root1.Left, root2.Right) && dfs(root1.Right, root2.Left)
  }
  return dfs(root, root)
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

const dfs = (root1: TreeNode | null, root2: TreeNode | null) => {
  if (root1 == root2) {
    return true;
  }
  if (root1 == null || root2 == null || root1.val != root2.val) {
    return false;
  }
  return dfs(root1.left, root2.right) && dfs(root1.right, root2.left);
};

function isSymmetric(root: TreeNode | null): boolean {
  return dfs(root.left, root.right);
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//   TreeNode {
//     val,
//     left: None,
//     right: None
//   }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
  fn dfs(root1: &Option<Rc<RefCell<TreeNode>>>, root2: &Option<Rc<RefCell<TreeNode>>>) -> bool {
    if root1.is_none() && root2.is_none() {
      return true;
    }
    if root1.is_none() || root2.is_none() {
      return false;
    }
    let node1 = root1.as_ref().unwrap().borrow();
    let node2 = root2.as_ref().unwrap().borrow();
    node1.val == node2.val &&
      Self::dfs(&node1.left, &node2.right) &&
      Self::dfs(&node1.right, &node2.left)
  }

  pub fn is_symmetric(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
    let node = root.as_ref().unwrap().borrow();
    Self::dfs(&node.left, &node.right)
  }
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.left = (left===undefined ? null : left)
 *   this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function (root) {
  function dfs(root1, root2) {
    if (!root1 && !root2) return true;
    if (!root1 || !root2 || root1.val != root2.val) return false;
    return dfs(root1.left, root2.right) && dfs(root1.right, root2.left);
  }
  return dfs(root, root);
};

方法二

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//   TreeNode {
//     val,
//     left: None,
//     right: None
//   }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;
impl Solution {
  pub fn is_symmetric(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
    let root = root.unwrap();
    let mut node = root.as_ref().borrow_mut();
    let mut queue = VecDeque::new();
    queue.push_back([node.left.take(), node.right.take()]);
    while let Some([root1, root2]) = queue.pop_front() {
      if root1.is_none() && root2.is_none() {
        continue;
      }
      if root1.is_none() || root2.is_none() {
        return false;
      }
      if let (Some(node1), Some(node2)) = (root1, root2) {
        let mut node1 = node1.as_ref().borrow_mut();
        let mut node2 = node2.as_ref().borrow_mut();
        if node1.val != node2.val {
          return false;
        }
        queue.push_back([node1.left.take(), node2.right.take()]);
        queue.push_back([node1.right.take(), node2.left.take()]);
      }
    }
    true
  }
}