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101. Symmetric Tree

Description

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

 

Example 1:

Input: root = [1,2,2,3,4,4,3]
Output: true

Example 2:

Input: root = [1,2,2,null,3,null,3]
Output: false

 

Constraints:

  • The number of nodes in the tree is in the range [1, 1000].
  • -100 <= Node.val <= 100

 

Follow up: Could you solve it both recursively and iteratively?

Solutions

Solution 1: Recursion

We design a function $dfs(root1, root2)$ to determine whether two binary trees are symmetric. The answer is $dfs(root, root)$.

The logic of the function $dfs(root1, root2)$ is as follows:

  • If both $root1$ and $root2$ are null, then the two binary trees are symmetric, return true.
  • If only one of $root1$ and $root2$ is null, or if $root1.val \neq root2.val$, then the two binary trees are not symmetric, return false.
  • Otherwise, determine whether the left subtree of $root1$ is symmetric to the right subtree of $root2$, and whether the right subtree of $root1$ is symmetric to the left subtree of $root2$. Here we use recursion.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

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# 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)

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/**
 * 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);
    }
}

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/**
 * 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);
    }
};

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/**
 * 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)
}

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/**
 * 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