直径问题 Diameter Problems

2019-11-03 21:37:59

一、Diameter of Binary Tree

问题描述：

问题求解：

解法一、第一反应是树上动归，每个节点保存一下左右的最大深度，最后以每个节点作为中枢计算最大的长度即可。

    public int diameterOfBinaryTree(TreeNode root) {
        Map<TreeNode, int[]> map = new HashMap<>();
        dfs(root, map);
        int res = 0;
        for (TreeNode key : map.keySet()) {
            res = Math.max(res, map.get(key)[0] + map.get(key)[1]);
        }
        return res;
    }
    
    private int[] dfs(TreeNode root, Map<TreeNode, int[]> map) {
        if (root == null) return new int[]{-1, -1};
        int[] l = dfs(root.left, map);
        int[] r = dfs(root.right, map);
        map.put(root, new int[]{Math.max(l[0], l[1]) + 1, Math.max(r[0], r[1]) + 1});
        return map.get(root);
    }

解法二、不求直径，而是转求每个节点的最大深度，遍历的时候可以顺便得到直径。

    public int diameterOfBinaryTree(TreeNode root) {
        int[] res = new int[1];
        helper(root, res);
        return res[0];
    }
    
    private int helper(TreeNode root, int[] res) {
        if (root == null) return 0;
        int l = helper(root.left, res);
        int r = helper(root.right, res);
        res[0] = Math.max(res[0], l + r);
        return Math.max(l, r) + 1;
    }

 

二、Tree Diameter

问题描述：

问题求解： 

    int res = 0;
    public int treeDiameter(int[][] edges) {
        Map<Integer, Set<Integer>> graph = new HashMap<>();
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            if (!graph.containsKey(u)) graph.put(u, new HashSet<>());
            if (!graph.containsKey(v)) graph.put(v, new HashSet<>());
            graph.get(u).add(v);
            graph.get(v).add(u);
        }
        helper(graph, 0, new HashSet<Integer>());
        return res;
    }

    private int helper(Map<Integer, Set<Integer>> graph, int node, Set<Integer> used) {
        Integer first = null;
        Integer second = null;
        used.add(node);
        for (int next : graph.get(node)) {
            if (used.contains(next)) continue;
            int tmp = helper(graph, next, used);
            if (first == null || first < tmp) {
                second = first;
                first = tmp;
            } else if (second == null || second < tmp) second = tmp;
        }
        int l1 = first == null ? 0 : first + 1;
        int l2 = second == null ? 0 : second + 1;
        res = Math.max(res, l1 + l2);
        return l1;
    }