binary-tree/traversal/symmetric-tree

Symmetric Tree

描述

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

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

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

But the following [1,2,2,null,3,null,3] is not:

    1
   / \
  2   2
   \   \
   3    3

Note: Bonus points if you could solve it both recursively and iteratively.

分析

无

递归版

// Symmetric Tree
// 递归版，时间复杂度O(n)，空间复杂度O(logn)
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        return isSymmetric(root.left, root.right);
    }
    private static boolean isSymmetric(TreeNode p, TreeNode q) {
        if (p == null && q == null) return true;   // 终止条件
        if (p == null || q == null) return false;  // 终止条件
        return p.val == q.val      // 三方合并
                && isSymmetric(p.left, q.right)
                && isSymmetric(p.right, q.left);
    }
}

// Symmetric Tree
// 递归版，时间复杂度O(n)，空间复杂度O(logn)
class Solution {
public:
    bool isSymmetric(TreeNode *root) {
        if (root == nullptr) return true;
        return isSymmetric(root->left, root->right);
    }
    bool isSymmetric(TreeNode *p, TreeNode *q) {
        if (p == nullptr && q == nullptr) return true;   // 终止条件
        if (p == nullptr || q == nullptr) return false;  // 终止条件
        return p->val == q->val      // 三方合并
                && isSymmetric(p->left, q->right)
                && isSymmetric(p->right, q->left);
    }
};

迭代版

// Symmetric Tree
// 迭代版，时间复杂度O(n)，空间复杂度O(logn)
public class Solution {
    public boolean isSymmetric (TreeNode root) {
        if (root == null) return true;

        Stack<TreeNode> s = new Stack<>();
        s.push(root.left);
        s.push(root.right);

        while (!s.isEmpty()) {
            TreeNode p = s.pop ();
            TreeNode q = s.pop ();

            if (p == null && q == null) continue;
            if (p == null || q == null) return false;
            if (p.val != q.val) return false;

            s.push(p.left);
            s.push(q.right);

            s.push(p.right);
            s.push(q.left);
        }

        return true;
    }
}

// Symmetric Tree
// 迭代版，时间复杂度O(n)，空间复杂度O(logn)
class Solution {
public:
    bool isSymmetric (TreeNode* root) {
        if (!root) return true;

        stack<TreeNode*> s;
        s.push(root->left);
        s.push(root->right);

        while (!s.empty ()) {
            auto p = s.top (); s.pop();
            auto q = s.top (); s.pop();

            if (!p && !q) continue;
            if (!p || !q) return false;
            if (p->val != q->val) return false;

            s.push(p->left);
            s.push(q->right);

            s.push(p->right);
            s.push(q->left);
        }

        return true;
    }
};

相关题目

• Same Tree