101. Symmetric Tree

最新推荐文章于 2019-04-15 20:39:00 发布

原创最新推荐文章于 2019-04-15 20:39:00 发布 · 142 阅读

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博客包含题目描述和方法思路两部分内容，方法思路有迭代（iterative）和递归（recursive）两种途径，聚焦信息技术领域的解题思路。

题目描述

在这里插入图片描述

方法思路

Approach1: iterative

class Solution{
    //Runtime: 5 ms, faster than 100.00%  
    //Memory Usage: 39 MB, less than 30.40% 
    public boolean isSymmetric(TreeNode root) {
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        if(root == null) return true;
        q.add(root.left);
        q.add(root.right);
        while(q.size() > 0){
            TreeNode left = q.poll(), right = q.poll();
            if(left== null && right == null) continue;
            if(left == null || right == null) return false;
            if(left.val != right.val) return false;
            q.add(left.left);
            q.add(right.right);
            q.add(left.right);
            q.add(right.left);            
        }
        return true;
    }
}

Approach2: recursive

class Solution{
    //Memory Usage: 39.5 MB, less than 15.45%
    //Runtime: 5 ms, faster than 100.00%
    public boolean isSymmetric(TreeNode root) {
    return isMirror(root, root);
}

public boolean isMirror(TreeNode t1, TreeNode t2) {
    if (t1 == null && t2 == null) return true;
    if (t1 == null || t2 == null) return false;
    return (t1.val == t2.val)
        && isMirror(t1.right, t2.left)
        && isMirror(t1.left, t2.right);
}
}