Algorithm

101.symmetric-tree

Description:

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

Example:

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Input: root = [1,2,2,3,4,4,3]
         1
      /     \
     2        2
   /   \    /   \
  3     4  4     3

Output: true
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Input: root = [1,2,2,null,3,null,3]
         1
      /     \
     2        2
       \        \
        3        3

Output: false

Follow up: Could you solve it both recursively and iteratively?

Similar Question: 94.binary-tree-inorder-traversal

First Answer (Recursive):

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<?php

namespace LeetCode\SymmetricTree;

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     public $val = null;
 *     public $left = null;
 *     public $right = null;
 *     function __construct($val = 0, $left = null, $right = null) {
 *         $this->val = $val;
 *         $this->left = $left;
 *         $this->right = $right;
 *     }
 * }
 */
class Solution
{
    /**
     * @var bool
     */
    protected $answer = true;

    /**
     * @param TreeNode $root
     * @return bool
     */
    public function isSymmetric($root)
    {
        if (is_null($root->val)) {
            return true;
        }

        if (is_null($root->left) && is_null($root->right)) {
            return true;
        }

        $this->checkSysmmetric($root->left, $root->right);
        return $this->answer;
    }

    /**
     * @param TreeNode $leftTree
     * @param TreeNode $rightTree
     */
    private function checkSysmmetric($leftTree, $rightTree)
    {
        if ($leftTree->val !== $rightTree->val) {
            return $this->answer = false;
        }

        if (
            $leftTree->left instanceof TreeNode &&
            $rightTree->right instanceof TreeNode
        ) {
            $this->checkSysmmetric($leftTree->left, $rightTree->right);
        }

        if (
            $leftTree->right instanceof TreeNode &&
            $rightTree->left instanceof TreeNode
        ) {
            $this->checkSysmmetric($leftTree->right, $rightTree->left);
        }

        if ($this->isNullAmountOdd($leftTree->left, $rightTree->right)) {
            return $this->answer = false;
        }

        if ($this->isNullAmountOdd($leftTree->right, $rightTree->left)) {
            return $this->answer = false;
        }
    }

    /**
     * @param TreeNode|null $a
     * @param TreeNode|null $b
     * @return bool
     */
    private function isNullAmountOdd($a, $b)
    {
        $nullAmount = 0;

        if (is_null($a)) {
            $nullAmount++;
        }

        if (is_null($b)) {
            $nullAmount++;
        }

        return ($nullAmount % 2 == 1);
    }
}
  • Time Complexity: O(n)

Other Answer (Iterating + Stack):

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<?php

namespace LeetCode\SymmetricTree;

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     public $val = null;
 *     public $left = null;
 *     public $right = null;
 *     function __construct($val = 0, $left = null, $right = null) {
 *         $this->val = $val;
 *         $this->left = $left;
 *         $this->right = $right;
 *     }
 * }
 */
class Solution
{
    /**
     * @var array
     */
    protected $stack = [];

    /**
     * @param TreeNode $root
     * @return bool
     */
    public function isSymmetric($root)
    {
        if (is_null($root->val)) {
            return true;
        }

        $this->stack[] = [$root->left, $root->right];

        while (count($this->stack) > 0) {
            $pop = array_pop($this->stack);
            $left = $pop[0];
            $right = $pop[1];

            if (is_null($left) && is_null($right)) {
                continue;
            }

            if (is_null($left) || is_null($right)) {
                return false;
            }

            if ($left->val == $right->val) {
                $this->stack[] = [$left->left, $right->right];
                $this->stack[] = [$left->right, $right->left];
            } else {
                return false;
            }
        }

        return true;
    }
}
  • Time Complexity: O(n)

ʕ •ᴥ•ʔ：Iterating的作法，有體會到要利用Stack當存檔點，但具體的while條件沒有想出來。