[LeetCode] 101. Symmetric Tree

判断对称树。题意是给一个树，判断它是不是左右对称/轴对称的。例子

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

题目很简单，跟100题几乎一样，也是用DFS思路做。唯一不同的地方是当判断子树的时候，100题是判断left.left == right.left；而这题是判断left.right == right.left

时间O(n)

空间O(n)

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function(root) {
    if (root === null) return true;
    return helper(root.left, root.right);
};

var helper = 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 helper(left.left, right.right) && helper(left.right, right.left);
};

Java实现

class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null)
            return true;
        return helper(root.left, root.right);
    }

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

BFS解法

时间O(n)

空间O(n)

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function(root) {
    const q = [root, root];
    while (q.length) {
        const [l, r] = [q.shift(), q.shift()];
        if (!l && !r) continue;
        if (!!l !== !!r || l.val !== r.val) return false;
        q.push(l.left, r.right, l.right, r.left);
    }
    return true;
};