#leetcode#Symmetric Tree

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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);    }}

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