LeetCode刷题 Day15
226. Invert Binary Tree
Given the root of a binary tree, invert the tree, and return its root.
Example 1:
Input: root = [4,2,7,1,3,6,9]
Output: [4,7,2,9,6,3,1]
Example 2:
Input: root = [2,1,3]
Output: [2,3,1]
Example 3:
Input: root = []
Output: []
思路:
- 利用先序遍历交换节点 再逐层返回
代码:
var invertTree = function(root) {
if (!root) return null;
let temp = root.left;
root.left = root.right;
root.right = temp;
invertTree(root.left);
invertTree(root.right);
return root;
};
时间复杂度: O(n), 空间复杂度: O(n)
101. Symmetric Tree
Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
Example 1:
Input: root = [1,2,2,3,4,4,3]
Output: true
Example 2:
Input: root = [1,2,2,null,3,null,3]
Output: false
思路:
- 使用辅助函数，参数为root.left, root.right
- 为了避免null error, 需要做空节点判断 (!left && !right) return true; (!left || !right) return false;
代码:
var isSymmetric = function(root) {
if (!root) return true;
var helper = function(left, right) {
if (!left && !right) return true;
if (!left || !right) return false;
return left.val === right.val &&
helper(left.left, right.right) &&
helper(left.right, right.left);
}
return helper(root.left, root.right);
};
时间复杂度: O(n), 空间复杂度: O(n)
