leetcode 101. Symmetric Tree

最新推荐文章于 2020-03-19 16:43:39 发布

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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实现方式，并详细解释了每种方法的工作原理。

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:

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

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

用了递归的方法，每次比较对应位置的节点

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root==null) return true;
        return isMirror(root.left, root.right);
    }
    
    private boolean isMirror(TreeNode left, TreeNode right){
        if(left==null && right==null) return true;
        if(left==null || right==null) return false;
        
        if(left.val!=right.val) return false;
        return isMirror(left.left, right.right) && isMirror(left.right, right.left);
        
    }
}

附上迭代做法

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() > 1){
            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;
    }