[LeetCode-109] Symmetric Tree

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#leetcode #二叉树 #递归

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本文介绍了一种检查二叉树是否为中心对称的方法，提供了递归和迭代两种解决方案，并详细解释了每一步的判断逻辑。通过示例对比帮助理解二叉树对称性的概念。

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

For example, this binary tree is symmetric:

But the following is not:

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.

Analysis:

recursion

situations that left child is same with right child

left.val == right.val or left == right == null

left.left child == right.right child

left.right child == right.left child

java

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

c++

bool isSymmetric(TreeNode *root) {
        if(root == NULL) return true;
        return symmetric(root->left,root->right);
    }
    bool symmetric(TreeNode *left, TreeNode *right){
        if(left == NULL) return right==NULL;
        if(right == NULL) return left==NULL;
        if(left->val!=right->val) return false;
        if(!symmetric(left->left,right->right)) return false;
        if(!symmetric(left->right,right->left)) return false;
        return true;
    }