742. Closest Leaf in a Binary Tree查找最近的叶子节点

［抄题］：

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]：

时间分析：

空间分析：

 [优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

［思惟问题］：

不知道还要找点，把路径存在hashmap中

［英文数据结构或算法，为何不用别的数据结构或算法］：

［一句话思路］：

bfs的过程当中，cur节点的左、右、map中存储的路径都要放进q

［输入量］：空： 正常状况：特大：特小：程序里处理到的特殊状况：异常状况（不合法不合理的输入）：

［画图］：

［一刷］：

1. dfs函数的做用是返回有效的 值为k的节点，因此结果是左右节点的时候也须要返回
2. 左右节点均为空的时候，再返回root.val的数值

［二刷］：

1. dfs函数中，先把左节点放进去，再返回整个的left

［三刷］：

［四刷］：

［五刷］：

  [五分钟肉眼debug的结果]：

［总结］：

用map存路径map.put(root.left, root);，而后用dfs找到k

［复杂度］：Time complexity: O(n) Space complexity: O(n)

［算法思想：迭代/递归/分治/贪心］：

［关键模板化代码］：

［其余解法］：

［Follow Up］：

［LC给出的题目变变变］：

 [代码风格] ：

 [是否头一次写此类driver funcion的代码] ：

 [潜台词] ：

 

/** * 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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