力扣labuladong——一刷day62

提示：文章写完后，目录可以自动生成，如何生成可参考右边的帮助文档

---

前言

---

二叉树的递归分为「遍历」和「分解问题」两种思维模式，这道题需要用到「分解问题」的思维，而且要用到后序位置的妙用。

一、力扣1339. 分裂二叉树的最大乘积

/**
 * 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 {
    HashSet<Long> set = new HashSet<>();
    public int maxProduct(TreeNode root) {
        long sum = fun(root);
        double res = 0;
        double t = 1e9 + 7;
        for(long s : set){
            double temp = s * (sum - s);
            res = Math.max(res,temp);
        }
        return (int)(res%t);
    }
    public long fun(TreeNode root){
        if(root == null){
            return 0;
        }
        long left = fun(root.left);
        long right = fun(root.right);
        long cur = left + right + root.val;
        set.add(cur);
        return cur;
    }
}

二、力扣2049. 统计最高分的节点数目

class Solution {
    int[][] tree;
    HashMap<Long,Integer> scoreToCount = new HashMap<>();
    public int countHighestScoreNodes(int[] parents) {
        this.tree = buildTree(parents);
        countNode(0);
        long maxScore = 0;
        for(long score : scoreToCount.keySet()){
            maxScore = Math.max(maxScore,score);
        }
        return scoreToCount.get(maxScore);
    }

    int countNode(int root){
        if(root == -1){
            return 0;
        }
        int n = tree.length;
        int leftCount = countNode(tree[root][0]);
        int rightCount = countNode(tree[root][1]);
        int otherCount = n - leftCount - rightCount -1;
        long score = (long)Math.max(leftCount,1) * Math.max(rightCount,1) * Math.max(otherCount,1);
        scoreToCount.put(score,scoreToCount.getOrDefault(score,0)+1);
        return leftCount + rightCount + 1;
    }

    public int[][] buildTree(int[] parents){
        int n = parents.length;
        int[][] tree = new int[n][2];
        for(int i = 0; i < n ; i ++){
            tree[i][0] = tree[i][1] = -1;
        }
        for(int i = 1; i < n ; i ++){
            int parent_i = parents[i];
            if(tree[parent_i][0] == -1){
                tree[parent_i][0] = i;
            }else{
                tree[parent_i][1] = i;
            }
        }
        return tree;
    }
}

三、力扣1372. 二叉树中的最长交错路径

/**
 * 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 longestZigZag(TreeNode root) {
        fun(root);
        return res;
    }
    public int[] fun(TreeNode root){
        if(root == null){
            return new int[]{-1,-1};
        }
        int[] left = fun(root.left);
        int[] right = fun(root.right);
        int p1 = left[1] + 1;
        int p2 = right[0] + 1;
        res = Math.max(res,Math.max(p1,p2));
        return new int[]{p1,p2};
    }
}