Symmetric Tree

Description

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

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

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

But the following [1,2,2,null,3,null,3] is not:

    1
   / \
  2   2
   \   \
   3    3

Note:

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

Tags: Tree, Depth-first Search, Breadth-first Search

思路 0

题意是判断一棵二叉树是否左右对称，首先想到的是深搜，比较根结点的左右两棵子树是否对称，如果左右子树的值相同，那么再分别对左子树的左节点和右子树的右节点，左子树的右节点和右子树的左节点做比较即可。

java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        return root == null || helper(root.left, root.right);
    }

    public boolean helper(TreeNode left, TreeNode right) {
        if (left == null || right == null) return left == right;
        if (left.val != right.val) return false;
        return helper(left.left, right.right) && helper(left.right, right.left);
    }
}

kotlin(208ms/100.00%):

class Solution {

    private fun symmetric(left: TreeNode?, right: TreeNode?): Boolean {
        if (left == null && right == null) return true
        if (left == null || right == null) return false
        return left.`val` == right.`val` && symmetric(left.right, right.left) && symmetric(left.left, right.right)
    }

    fun isSymmetric(root: TreeNode?): Boolean {
        if (root == null) return true
        if (root.left == null && root.right == null) return true
        if (root.left == null || root.right == null) return false
        return symmetric(root.left, root.right)
    }
}

JavaScript：

var isSymmetric = function(root) {
    if(root == null) return true
    var x = function(left, right) {
        if(left == null && right == null) return true
        if(left == null || right == null) return false
        if (left.val !== right.val) return false;
        return x(left.left, right.right) && x(left.right, right.left) 
    }
    return x(root.left, root.right)
};

思路 1

第二种思路就是宽搜了，宽搜肯定要用到队列，Java 中可用 LinkedList 替代，也是要做到左子树的左节点和右子树的右节点，左子树的右节点和右子树的左节点做比较即可。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
     public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        LinkedList<TreeNode> q = new LinkedList<>();
        q.add(root.left);
        q.add(root.right);
        TreeNode left, right;
        while (q.size() > 1) {
            left = q.pop();
            right = q.pop();
            if (left == null && right == null) continue;
            if (left == null || right == null) return false;
            if (left.val != right.val) return false;
            q.add(left.left);
            q.add(right.right);
            q.add(left.right);
            q.add(right.left);
        }
        return true;
    }
}

结语

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