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[ganu] Week 10 #1008
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[ganu] Week 10 #1008
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2ffc9a0
feat: 226. Invert Binary Tree
gwbaik9717 322b54d
feat: 33. Search in Rotated Sorted Array
gwbaik9717 aca3c7d
feat: 55. Jump Game
gwbaik9717 a69c944
feat: 207. Course Schedule
gwbaik9717 27fac22
feat: 23. Merge k Sorted Lists
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Original file line number | Diff line number | Diff line change |
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// Time complexity: O(n) | ||
// Space complexity: O(n) | ||
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/** | ||
* Definition for a binary tree node. | ||
* function TreeNode(val, left, right) { | ||
* this.val = (val===undefined ? 0 : val) | ||
* this.left = (left===undefined ? null : left) | ||
* this.right = (right===undefined ? null : right) | ||
* } | ||
*/ | ||
/** | ||
* @param {TreeNode} root | ||
* @return {TreeNode} | ||
*/ | ||
var invertTree = function (root) { | ||
const dfs = (current) => { | ||
if (!current) { | ||
return; | ||
} | ||
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const temp = current.left; | ||
current.left = current.right; | ||
current.right = temp; | ||
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dfs(current.left); | ||
dfs(current.right); | ||
}; | ||
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dfs(root); | ||
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return root; | ||
}; | ||
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DFS와 재귀를 활용하면 더 깔끔한 코드로 구현할 수 있겠어요. 참고가 되었습니다. 감사합니다!