LeetCode101. Symmetric Tree

101.Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
给一个二叉树，判断它是否是中心对称二叉树。
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.

public class TreeNode {         int val;    TreeNode left;    TreeNode right;    TreeNode(int x) { val = x; }}

因为刚做了 LeetCode.NO100 SameTree ，所以一下就想到了用相似的方法，用栈的方法，只是遍历顺序不一样，将二叉树从根开始，左右分成两棵子树，然后一棵先遍历左边，另一棵先遍历右边即可。

public boolean isSymmetric(TreeNode root) {    if(root == null) return true;    if(root.left == null && root.right == null) return true;    if(root.left == null || root.right == null) return false;    TreeNode leftTree = root.left;    TreeNode rightTree = root.right;    Stack<TreeNode> stack_left = new Stack<>();    Stack<TreeNode> stack_right = new Stack<>();            stack_left.push(leftTree);    stack_right.push(rightTree);    while(!stack_left.isEmpty() && !stack_right.isEmpty()){        TreeNode left = stack_left.pop();        TreeNode right = stack_right.pop();        if(left.val != right.val) return false;        if(left.left != null) stack_left.push(left.left);        if(right.right != null) stack_right.push(right.right);        if(stack_left.size() != stack_right.size()) return false;        if(left.right != null) stack_left.push(left.right);        if(right.left != null) stack_right.push(right.left);        if(stack_left.size() != stack_right.size()) return false;    }           return stack_left.size() == stack_right.size();}

方法二：采用递归方法

public boolean isSymmetric1(TreeNode root){    if(root == null) return true;    else{        return isMirror1(root.left, root.right);    }}public boolean isMirror1(TreeNode left, TreeNode right){    if(left == null && right == null) return true;    if(left == null || right == null) return false;    if(left.val == right.val){        return isMirror1(left.left, right.right) && isMirror1(left.right, right.left);    }    return false;}

方法三：
跟上面方法不同就是第一次调用 isMirror 方法，isMirror2(root, root)，上面的方法是判断两棵树是否是镜像树，
下面这个判断一棵树是否是镜像树。
分析时间复杂度：每个节点都会对比一次，所以总共运行次数是 O(n),n是树的节点个数。
函数递归调用次数与树的高度有关，最坏的情况，树是线性树，高度是n，因此栈的空间复杂度最坏是O(n)。

public boolean isSymmetric2(TreeNode root) {    return isMirror2(root, root);}public boolean isMirror2(TreeNode t1, TreeNode t2) {    if (t1 == null && t2 == null) return true;    if (t1 == null || t2 == null) return false;    return (t1.val == t2.val)      && isMirror2(t1.right, t2.left)      && isMirror2(t1.left, t2.right);}

方法四：用队列的方法
用队列的方法：比较队列中相邻两个元素结构、值是否相等。
分析时间复杂度：总共运行次数是O(n)，n是树的节点个数。
空间复杂度：O(n)

public boolean isSymmetric3(TreeNode root) {    Queue<TreeNode> q = new LinkedList<>();    q.add(root);    q.add(root);    while (!q.isEmpty()) {        TreeNode t1 = q.poll();        TreeNode t2 = q.poll();        if (t1 == null && t2 == null) continue;        if (t1 == null || t2 == null) return false;        if (t1.val != t2.val) return false;        q.add(t1.left);        q.add(t2.right);        q.add(t1.right);        q.add(t2.left);    }    return true;}