742. Closest Leaf in a Binary Tree查找最近的葉子節點

［抄題］：

Given a binary tree where every node has a unique value, and a target key k, find the value of the nearest leaf node to target k in the tree.

Here, nearest to a leaf means the least number of edges travelled on the binary tree to reach any leaf of the tree. Also, a node is called a leaf if it has no children.

In the following examples, the input tree is represented in flattened form row by row. The actual root tree given will be a TreeNode object.

Example 1:

Input: root = [1, 3, 2], k = 1 Diagram of binary tree: 1 / \ 3 2 Output: 2 (or 3) Explanation: Either 2 or 3 is the nearest leaf node to the target of 1. 

 

Example 2:

Input: root = [1], k = 1 Output: 1 Explanation: The nearest leaf node is the root node itself. 

 

Example 3:

Input: root = [1,2,3,4,null,null,null,5,null,6], k = 2 Diagram of binary tree: 1 / \ 2 3 / 4 / 5 / 6 Output: 3 Explanation: The leaf node with value 3 (and not the leaf node with value 6) is nearest to the node with value 2.

 [暴力解法]：

時間分析：

空間分析：

 [優化後]：

時間分析：

空間分析：

［奇葩輸出條件］：

［奇葩corner case］：

［思惟問題］：

不知道還要找點，把路徑存在hashmap中

［英文數據結構或算法，爲何不用別的數據結構或算法］：

［一句話思路］：

bfs的過程當中，cur節點的左、右、map中存儲的路徑都要放進q

［輸入量］：空： 正常狀況：特大：特小：程序裏處理到的特殊狀況：異常狀況（不合法不合理的輸入）：

［畫圖］：

［一刷］：

1. dfs函數的做用是返回有效的 值爲k的節點，因此結果是左右節點的時候也須要返回
2. 左右節點均爲空的時候，再返回root.val的數值

［二刷］：

1. dfs函數中，先把左節點放進去，再返回整個的left

［三刷］：

［四刷］：

［五刷］：

  [五分鐘肉眼debug的結果]：

［總結］：

用map存路徑map.put(root.left, root);，而後用dfs找到k

［複雜度］：Time complexity: O(n) Space complexity: O(n)

［算法思想：迭代/遞歸/分治/貪心］：

［關鍵模板化代碼］：

［其餘解法］：

［Follow Up］：

［LC給出的題目變變變］：

 [代碼風格] ：

 [是否頭一次寫此類driver funcion的代碼] ：

 [潛臺詞] ：

 

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */
class Solution { public int findClosestLeaf(TreeNode root, int k) { //corner case
        if (root == null) return -1; //initialiazation: map, q //put first node into q, add left, right, route
        HashMap<TreeNode, TreeNode> map = new HashMap<TreeNode, TreeNode>(); PriorityQueue<TreeNode> q = new PriorityQueue<TreeNode>(); TreeNode match = dfsTree(k, root, map); q.add(match); while (!q.isEmpty()) { TreeNode cur = q.poll(); if (cur.left == null && cur.right == null) return cur.val; if (cur.left != null) q.add(cur.left); if (cur.right != null) q.add(cur.right); if (map.containsKey(cur)) { q.add(map.get(cur)); map.remove(cur); } } return -1; } public TreeNode dfsTree(int k, TreeNode root, Map<TreeNode, TreeNode> map) { //corner case : null
        if (root == null) return null; //return if left & right is null
        if (root.val == k) return root; //put left into map and return left
        if (root.left != null) { map.put(root.left, root); TreeNode left = dfsTree(k, root.left, map); if (left != null) return left; } //put left into map and return left
        if (root.right != null) { map.put(root.right, root); TreeNode right = dfsTree(k, root.right, map); if (right != null) return right; } //return null
        return null; } }

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