[leetCode]968. 监控二叉树

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {

    // 每个节点右三种状态:0.无覆盖 1.有摄像头 2.有覆盖
    private final Integer UNCOVERED = 0, COVERED = 2, HASCAM = 1; 

    private Integer result = 0;

    public int minCameraCover(TreeNode root) {
        // 4. 递归结束如果根节点无覆盖则需要放摄像头
        if (UNCOVERED == traversal(root)) {
            result++;
        }
        return result;
    }

    private int traversal(TreeNode cur) {
        // 空节点,该节点有覆盖
        if (cur ==  null) return COVERED;
        int left = traversal(cur.left); // 左
        int right = traversal(cur.right); // 右

        // 逻辑处理     中
        // 1.如果左子节点和右子节点都覆盖了那么当前节点为无覆盖                  
        if (left == COVERED && right == COVERED) 
            return 0;
        // 2.有一个孩子没覆盖,父节点就应该放摄像头
        if (left == 0 || right == 0) {
            result++;
            return HASCAM; 
        }
        // 3.如果左右节点只要其中一个有摄像头则该节点为覆盖状态
        if (left == 1 || right == 1) {
            return COVERED;
        }
        return -1;
    }
}
原文地址:https://www.cnblogs.com/PythonFCG/p/13859882.html