leetcode101. 对称二叉树

1.题目描述：

给你一个二叉树的根节点root，检查它是否轴对称。

2.翻转树的递归解法：

自己在做这道题目的时候，发现递归第二步，每级递归需要做什么不好写，于是想反转右子树后再判断与左子树是否一致即可，翻转的题目正好做过leetcode226. 翻转二叉树。这种方法时间复杂度近似O(n)，空间复杂度最差O(n)。直接看代码：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return false;
        return symmetricTree(root.left,reverseTree(root.right));
    }
    public TreeNode reverseTree(TreeNode root){
        if(root == null) return null;
        TreeNode l = root.left;
        root.left = root.right;
        root.right = l;
        reverseTree(root.left);
        reverseTree(root.right);
        return root;
    }
    public boolean symmetricTree(TreeNode node1,TreeNode node2){
        if(node1 == null && node2 == null) return true;
        if(node1 == null || node2 == null) return false;
        return node1.val == node2.val && symmetricTree(node1.left,node2.left) && symmetricTree(node1.right,node2.right);
    }
}

 3.普通的递归解法：借助两次root，递归判断p左子树与q右子树是否镜像，p右子树与q左子树是否镜像。时间复杂度近似O(n)，空间复杂度链表最差O(n)，代码如下：

class Solution {
    public boolean isSymmetric(TreeNode root) {
        return check(root, root);
    }
    public boolean check(TreeNode p, TreeNode q) {
        if(p == null && q == null) return true;
        if(p == null || q == null) return false;
        return p.val == q.val && check(p.left,q.right) && check(p.right,q.left);
    }
}

4.在此基础上，BFS解法：

class Solution {
    public boolean isSymmetric(TreeNode root) {
        return check(root, root);
    }
    public boolean check(TreeNode u, TreeNode v) {
        Queue q = new LinkedList();
        q.offer(u);
        q.offer(v);
        while(!q.isEmpty()){
            u = q.poll();
            v = q.poll();
            if(u == null && v == null) continue;//重点注意
            if((u == null || v == null) || (u.val != v.val)) return false;
            q.offer(u.left);
            q.offer(v.right);
            q.offer(u.right);
            q.offer(v.left);
        }
        return true;
    }
}

5.二刷递归：递归比较左边的左边和右边的左边，左边的右边和右边的左边即可。

class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        return dfs(root.left, root.right);
    }
    public boolean dfs(TreeNode left, TreeNode right) {
        if (left == null && right == null) return true;
        if (left == null || right == null || left.val != right.val) return false;
        return dfs(left.left, right.right) && dfs(left.right, right.left);
    }
}

迭代：与层序遍历顺序不同，队列里一次放两个对称性的位置在同时弹出比较。

class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        Queue queue = new LinkedList<>();
        queue.offer(root.left);
        queue.offer(root.right);
        while (!queue.isEmpty()) {
            TreeNode n1 = queue.poll();
            TreeNode n2 = queue.poll();
            if (n1 == null && n2 == null) continue;
            if (n1 == null || n2 == null || n1.val != n2.val) return false;//判断逻辑与递归一致
            queue.offer(n1.left);//与普通层序遍历不同（不用考虑是否为null）
            queue.offer(n2.right);
            queue.offer(n1.right);
            queue.offer(n2.left);
        }
        return true;
    }
}