20190515

Author: mJackie

code/lc559.java

@@ -0,0 +1,36 @@
+package code;
+
+import java.util.List;
+
+/*
+ * 559. Maximum Depth of N-ary Tree
+ * 题意：多叉树的最大深度
+ * 难度：Easy
+ * 分类：Tree, Depth-first Search, Breadth-first Search
+ * 思路：BFS 也能做和 lc104思路一样
+ * Tips：lc104, lc111
+ */
+
+public class lc559 {
+    class Node {
+        public int val;
+        public List<Node> children;
+
+        public Node() {}
+
+        public Node(int _val,List<Node> _children) {
+            val = _val;
+            children = _children;
+        }
+    }
+
+    public int maxDepth(Node root) {
+        if(root==null) return 0;
+        int res = 0;
+        for (Node nd : root.children) {
+            res = Math.max(maxDepth(nd), res);
+        }
+        return res+1;
+    }
+
+}

code/lc589.java

@@ -0,0 +1,41 @@
+package code;
+
+import java.util.ArrayList;
+import java.util.List;
+
+/*
+ * 589. N-ary Tree Preorder Traversal
+ * 题意：多叉树先序遍历
+ * 难度：Easy
+ * 分类：Tree
+ * 思路：
+ * Tips：
+ */
+public class lc589 {
+    class Node {
+        public int val;
+        public List<Node> children;
+
+        public Node() {}
+
+        public Node(int _val,List<Node> _children) {
+            val = _val;
+            children = _children;
+        }
+    }
+
+    List<Integer> res;
+    public List<Integer> preorder(Node root) {
+        res = new ArrayList<>();
+        helper(root);
+        return res;
+    }
+
+    public void helper(Node root){
+        if(root==null) return;
+        res.add(root.val);
+        for(Node nd:root.children){
+            helper(nd);
+        }
+    }
+}

code/lc590.java

@@ -0,0 +1,33 @@
+package code;
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class lc590 {
+    class Node {
+        public int val;
+        public List<Node> children;
+
+        public Node() {}
+
+        public Node(int _val,List<Node> _children) {
+            val = _val;
+            children = _children;
+        }
+    }
+
+    List<Integer> res;
+    public List<Integer> postorder(Node root) {
+        res = new ArrayList<>();
+        helper(root);
+        return res;
+    }
+
+    public void helper(Node root){
+        if(root==null) return;
+        for(Node nd:root.children){
+            helper(nd);
+        }
+        res.add(root.val);
+    }
+}

code/lc834.java

@@ -0,0 +1,64 @@
+package code;
+
+import java.util.HashMap;
+import java.util.HashSet;
+/*
+ * 834. Sum of Distances in Tree
+ * 题意：求树中每个节点的值，到其他节点的距离和
+ * 难度：Hard
+ * 分类：Tree, Depth-first Search
+ * 思路：真的难，我是做不出来
+ *      两次遍历，第一次后续遍历，计算出每个节点为根的树，有几个孩子节点，并计算出root的结果
+ *      根据前一步的计算结果，开始先序遍历，一步步计算出其他节点的结果
+ *      res[node] = res[parent]-2*count[node]+count.length;
+ *      https://leetcode.com/problems/sum-of-distances-in-tree/discuss/130583/C%2B%2BJavaPython-Pre-order-and-Post-order-DFS-O(N)
+ *      https://leetcode.com/problems/sum-of-distances-in-tree/discuss/130567/Two-traversals-O(N)-python-solution-with-Explanation
+ * Tips：
+ */
+public class lc834 {
+    public static void main(String[] args) {
+        int[][] arr = {{0,1},{0,2},{2,3},{2,4},{2,5}};
+        sumOfDistancesInTree(6, arr);
+    }
+
+    static int[] count;
+    static int[] res;
+    static HashMap<Integer, HashSet<Integer>> hm = new HashMap();
+    public static int[] sumOfDistancesInTree(int N, int[][] edges) {
+        if(edges.length==0) return new int[]{0};
+        count = new int[N]; //记录n为跟的树，下边有多少个节点
+        res = new int[N];
+        for (int i = 0; i < edges.length ; i++) {   //双向都添加，遍历的时候判断一下，因为并不一定0就是根
+            HashSet<Integer> hs = hm.getOrDefault(edges[i][0], new HashSet());
+            hs.add(edges[i][1]);
+            hm.put(edges[i][0], hs);
+            hs = hm.getOrDefault(edges[i][1], new HashSet());
+            hs.add(edges[i][0]);
+            hm.put(edges[i][1], hs);
+        }
+        helper1(0, -1);
+        helper2(0, -1);
+        return res;
+    }
+
+    public static int helper1(int node, int parent){ //后序遍历，求root对应的结果，计算每个节点的count
+        HashSet<Integer> hs = hm.getOrDefault(node, new HashSet());
+        for (Integer i:hs) {
+            if(i==parent) continue;     //是parent的话直接略过
+            count[node] += helper1(i, node);
+            res[node] += res[i] + count[i];
+        }
+        count[node]++;
+        return count[node];
+    }
+
+    public static void helper2(int node, int parent){   //先序遍历，求结果
+        HashSet<Integer> hs = hm.getOrDefault(node, new HashSet());
+        if(node!=0)
+            res[node] = res[parent]-2*count[node]+count.length; //转移计算
+        for (Integer i:hs) {
+            if(i==parent) continue; //是parent的话直接略过
+            helper2(i, node);
+        }
+    }
+}

readme.md

@@ -212,16 +212,20 @@ LeetCode 指南
 | 494 [Java](./code/lc494.java)
 | 538 [Java](./code/lc538.java)
 | 543 [Java](./code/lc543.java)
+| 559 [Java](./code/lc559.java)
 | 560 [Java](./code/lc543.java)
 | 572 [Java](./code/lc572.java)
 | 581 [Java](./code/lc581.java)
+| 589 [Java](./code/lc589.java)
+| 590 [Java](./code/lc590.java)
 | 617 [Java](./code/lc617.java)
 | 621 [Java](./code/lc621.java)
 | 647 [Java](./code/lc647.java)
 | 685 [Java](./code/lc685.java)
 | 714 [Java](./code/lc714.java)
 | 746 [Java](./code/lc746.java)
 | 771 [Java](./code/lc771.java)
+| 834 [Java](./code/lc834.java)
 | 877 [Java](./code/lc877.java)
 | 921 [Java](./code/lc921.java)
 | 922 [Java](./code/lc922.java)