力扣labuladong——一刷day60

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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，而且涉及处理子树，需要用后序遍历。

一、力扣663. 均匀树划分

/**
 * 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<Integer> set = new HashSet<>();
    public boolean checkEqualTree(TreeNode root) {
        int count = root.val + fun(root.left) + fun(root.right);
        if(count % 2 != 0){
            return false;
        }
        return set.contains(count/2);
    }
    public int  fun(TreeNode root){
        if(root == null){
            return  0;
        }
        int left = fun(root.left);
        int right = fun(root.right);
        int cur = left + right + root.val;
        set.add(cur);
        return cur;
    }
}

二、力扣687. 最长同值路径

/**
 * 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 longestUnivaluePath(TreeNode root) {
        if(root == null){
            return 0;
        }
        fun(root);
        return res-1;
    }
    public int fun(TreeNode root){
        if(root == null){
            return 0;
        }
        int left = fun(root.left);
        int right = fun(root.right);
        int cur = 1, curLeft = 1, curRight = 1;
        if(root.left != null){
            if(root.left.val == root.val){
                cur += left;
                curLeft += left;
            }
        }
        if(root.right != null){
            if(root.right.val == root.val){
                cur += right;
                curRight += right;
            }
        }
        res = Math.max(res,cur);
        return Math.max(curLeft,curRight);
    }
}

三、力扣814. 二叉树剪枝

/**
 * 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 {
    boolean flag = true;
    public TreeNode pruneTree(TreeNode root) {
        if(root == null){
            return null;
        }
        fun(root);
        if(flag){
            return null;
        }
        flag = true;
        root.left = pruneTree(root.left);
        root.right = pruneTree(root.right);
        return root;
    }
    public void fun(TreeNode root){
        if(root == null){
            return ;
        }
        if(root.val == 1){
            flag = false;
        }
        fun(root.left);
        fun(root.right);
    }
}

减枝

/**
 * 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 {
    public TreeNode pruneTree(TreeNode root) {
        if(root == null){
            return null;
        }
        root.left = pruneTree(root.left);
        root.right = pruneTree(root.right);
        if(root.left == null && root.right == null && root.val == 0){
            return null;
        }
        return root;
    }
}