求二叉树中最多的黑色节点

Problem: given a tree, color nodes black as many as possible without coloring two adjacent nodes

 

思路: 如果根节点r被标记为黑色，则其直接子节点不能被标记为黑色，如果不标记r，则其子节点可以标记为黑色，得到状态转移方程:

find(r) = max(black(r), white(r))

black(r)为当节点r被标记为黑色时，以r根节点的树中最多可标记为黑色的节点数

white(r)为当节点r不被标记为黑色时，以r根节点的树中最多可标记为黑色的节点数

则

white(r) = find(r.left) + find(r.right)  

//当r没被标记为黑色时，对其左右子节点继续寻找

 

black(r) = 1 + whilte(r.left) + white(r.right)

//当r被标记为黑色时，它自身对总黑色节点数贡献1，然后它的左右子节点一定是非黑色节点

package com.self;

public class DP_03_Tree {

    public static void main(String[] args) {
        
        Node node4 = new Node(4);
        Node node3 = new Node(3);
        Node node5 = new Node(5);
        Node node2 = new Node(2);
        Node node1 = new Node(1);
        
        node4.left = node3;
        node4.right = node5;
        node3.left = node2;
        node3.right = node1;
        
        DP_03_Tree app = new DP_03_Tree();
        int maxBlack = app.find(node4);
        System.out.println(maxBlack);
    }
    
    int find(Node node){
        if(null == node) return 0;
        if(null == node.left && null == node.right) return 1;
        return Math.max(findFromWhite(node), findFromBlack(node));
    }
    
    //Current node is white
    int findFromWhite(Node node){
        if(null == node) return 0;
        return find(node.left)+find(node.right);
    }
    
    //Current node is black
    int findFromBlack(Node node){
        if(null == node) return 0;
        //Current node is black, then its children do not count, but itself counts
        return 1 + findFromWhite(node.left) + findFromWhite(node.right);
    }
}

class Node{
    int v;
    Node left;
    Node right;
    Node(int v){
        this.v = v;
    }
}