Symmetric Tree

Leetcode

題意：

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

For example, this binary tree is symmetric:

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

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

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

解題思路：

Iterative法：

/**
 * 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;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root.left);
        queue.offer(root.right);

        while(!queue.isEmpty()){
            TreeNode left = queue.poll();
            TreeNode right = queue.poll();
            if(left == null && right == null) continue;
            if(left == null || right == null) return false;
            if(left.val != right.val) return false;
            queue.offer(left.left);
            queue.offer(right.right);
            queue.offer(left.right);
            queue.offer(right.left);

        }
        return true;

    }
}
}

遞迴法：

public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        return isMirror(root.left, root.right);
    }

    private boolean isMirror(TreeNode p, TreeNode q) {
        if (p == null && q == null) {
            return true;
        } else if (p == null || q == null) {
            return false;
        }
        if (p.val != q.val) {
            return false;
        }

        return isMirror(p.left, q.right) && isMirror(p.right, q.left);
    }
}

非遞迴法：分別用que來作dfs，一棵先左再右，另一棵先右再左。

public class Solution {
    public boolean isSymmetric(TreeNode root) {
        Queue<TreeNode> ql = new LinkedList<TreeNode>();
        Queue<TreeNode> qr = new LinkedList<TreeNode>();

        if(root == null) return true;
        if(root.left != null)
            ql.offer(root.left);
        if(root.right != null)
            qr.offer(root.right);

        while(ql.size() != 0 && qr.size() != 0){
            TreeNode left = ql.poll();
            TreeNode right = qr.poll();
            if(left.val != right.val 
                || (left.left == null && right.right != null)
                || (left.left != null && right.right == null)
                || (left.right == null && right.left != null)
                || (left.right != null && right.left == null))
                return false;

            if(left.left != null) ql.offer(left.left);

            if(left.right != null) ql.offer(left.right);

            if(right.right != null) qr.offer(right.right);

            if(right.left != null) qr.offer(right.left);
        }

        return ql.size() == qr.size();
    }
}