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前言

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说过后序位置的特殊之处，后序位置可以接收到子树的信息，同时也可以通过函数参数接收到父节点传递的信息，这道题就可以比较完美地体现这一特点

一、力扣1245. 树的直径

class Solution {int max = 0;Map<Integer,List<Integer>> map = new HashMap<>();HashSet<Integer> set = new HashSet<>();public int treeDiameter(int[][] edges) {if(edges.length == 0){return 0;}for(int[] e : edges){int a = e[0], b = e[1];if(!map.containsKey(a)){map.put(a,new ArrayList<>());}if(!map.containsKey(b)){map.put(b, new ArrayList<>());}map.get(a).add(b);map.get(b).add(a);}fun(edges[0][0]);return max;}public int fun(int root){if(set.contains(root)){return 0;}set.add(root);int first = 0, second = 0;for(int a : map.get(root)){int child = fun(a);if(child >= first){second = first;first = child;}else if(child > second){second = child;}}int cur = first + second;max = Math.max(max, cur);return first + 1;}
}

二、力扣968. 监控二叉树

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {int res = 0;public int minCameraCover(TreeNode root) {if(fun(root) == 0){res ++;}return res;}//0表示未覆盖//1表示安装摄像头//2表示已覆盖public int fun(TreeNode root){if(root == null){return 2;}int left = fun(root.left);int right = fun(root.right);if(left == 0 || right == 0){res ++;return 1;}if(left == 1 || right == 1){return 2;}if(left == 2 && right == 2){return 0;}return 0;}
}

三、力扣979. 在二叉树中分配硬币

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {int res = 0;public int distributeCoins(TreeNode root) {fun(root);return res;}public int fun(TreeNode root){if(root == null){return 0;}int left = fun(root.left);int right = fun(root.right);res += Math.abs(left) + Math.abs(right) + root.val - 1;return left + right + root.val - 1;}
}