#leetcode#Symmetric Tree

标签：leetcode

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.

分析： 写代码的时候感觉有点绕，第一个节点root没什么好判断的，递归也好，迭代也好，真正把树分成两半去比较都是从root.left 和 root.right 开始的， 然后就是 root.left.left & root.right.right, root.left.right & root.right.left

Iteratively:

/**
 * 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;
        }
        LinkedList<TreeNode> queuel = new LinkedList<>();
        LinkedList<TreeNode> queuer = new LinkedList<>();
        if(root.left != null){
            queuel.offer(root.left);
        }
        if(root.right != null){
            queuer.offer(root.right);
        }
        if(queuel.size() != queuer.size()){
            return false;
        }
        while(!queuel.isEmpty() && !queuer.isEmpty()){
            int sizel = queuel.size();
            int sizer = queuer.size();
            if(sizel != sizer){
                return false;
            }
            for(int i = 0; i < sizel; i++){
                TreeNode l = queuel.poll();
                TreeNode r = queuer.poll();
                if(l.val != r.val){
                    return false;
                }
                if(l.left != null){
                    if(r.right == null){
                        return false;
                    }
                    queuel.offer(l.left);
                    queuer.offer(r.right);
                }else{
                    if(r.right != null){
                        return false;
                    }
                }
                if(l.right != null){
                    if(r.left == null){
                        return false;
                    }
                    queuel.offer(l.right);
                    queuer.offer(r.left);
                }else{
                    if(r.left != null){
                        return false;
                    }
                }
            }
        }
        return true;
    }
}

recursively:

/**
 * 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 isSymmetric(root.left, root.right);
    }
    
    private boolean isSymmetric(TreeNode root1, TreeNode root2){
        if(root1 == null && root2 == null){
            return true;
        }
        if(root1 == null || root2 == null){
            return false;
        }
        if(root1.val != root2.val){
            return false;
        }
        return isSymmetric(root1.left, root2.right) && isSymmetric(root1.right, root2.left);
    }
}

#leetcode#Symmetric Tree

标签：leetcode

原文地址：http://blog.csdn.net/chibaoneliuliuni/article/details/47003771