力扣labuladong——一刷day58

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#leetcode #算法 #职场和发展 #java #数据结构

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文章介绍了如何使用递归方法解决四道力扣（LeetCode）中的二叉树问题：1.找出最大BST子树；2.寻找二叉树的叶子节点；3.计算出现次数最多的子树元素之和；4.计算二叉树的坡度。每个问题都涉及到了二叉树遍历和问题分解技巧。

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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，这道题需要混用两种思维。

一、力扣333. 最大 BST 子树

/**
 * 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 count = 0;
    int res = 0;
    TreeNode pre = null;
    boolean flag = true;
    public int largestBSTSubtree(TreeNode root) {
        if(root == null){
            return 0;
        }
        fun(root);
        if(flag){
            res = Math.max(res,count);
            System.out.println(res);
        }
        flag = true;
        count = 0;
        pre = null;
        largestBSTSubtree(root.left);
        largestBSTSubtree(root.right);
        return res;
    }
    public void fun(TreeNode root){
        if(root == null){
            return ;
        }
        count ++;
        fun(root.left);
        if(pre != null){
            if(root.val <= pre.val){
                flag = false;
            }
        }
        pre = root;
        fun(root.right);
    }
}

后序

/**
 * 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 largestBSTSubtree(TreeNode root) {
        fun(root);
        return res;
    }
    //第一个为左子树最大值，第二个为右子树最小值，第三个为节点数
    public int[] fun(TreeNode root){
        if(root == null){
            return new int[]{Integer.MAX_VALUE,Integer.MIN_VALUE,0};
        }
        int[] leftArr = fun(root.left);
        int[] rightArr = fun(root.right);
        if(leftArr == null || rightArr == null){
            return null;
        }
        int leftMin = leftArr[0], leftMax = leftArr[1], leftCount = leftArr[2];
        int rightMin = rightArr[0], rightMax = rightArr[1], rightCount = rightArr[2];
        if(root.val > leftMax && root.val < rightMin){
            int curMin = Math.min(leftMin,root.val);
            int curMax = Math.max(rightMax,root.val);
            int curCount = leftCount + rightCount +1;
            res = Math.max(res,curCount);
            return new int[]{curMin, curMax, curCount};
        }
        return null;
    }
}

二、力扣366. 寻找二叉树的叶子节点

/**
 * 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 {
    List<List<Integer>> res = new ArrayList<>();
    public List<List<Integer>> findLeaves(TreeNode root) {
        fun(root);
        return res;
    }
    public int fun(TreeNode root){
        if(root == null){
            return 0;
        }
        int h = Math.max(fun(root.left),fun(root.right)) + 1;
        if(res.size() < h){
            res.add(new ArrayList<>());
        }
        res.get(h-1).add(root.val);
        return h;
    }
}

三、力扣508. 出现次数最多的子树元素和

/**
 * 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 flag = 0;
    Map<Integer,Integer> map = new HashMap<>();
    public int[] findFrequentTreeSum(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        fun(root);
        Set<Map.Entry<Integer,Integer>> entrySet = map.entrySet();
        entrySet.stream().forEach(entry -> {
            int key = entry.getKey();
            int value = entry.getValue();
            if(value == flag){
                list.add(key);
            }
        });
        return list.stream().mapToInt(i -> i).toArray();
    }
    public int fun(TreeNode root){
        if(root == null){
            return 0;
        }
        int leftSum = fun(root.left);
        int rightSum = fun(root.right);
        int curSum = leftSum + rightSum + root.val;
        map.put(curSum,map.getOrDefault(curSum,0)+1);
        flag = Math.max(flag,map.get(curSum));
        return curSum;
    }
}

四、力扣563. 二叉树的坡度

/**
 * 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 findTilt(TreeNode root) {
        fun(root);
        return res;
    }
    public int fun(TreeNode root){
        if(root == null){
            return 0;
        }
        int leftSum = fun(root.left);
        int rightSum = fun(root.right);
        int curK = Math.abs(leftSum - rightSum);
        res += curK;
        return (leftSum + rightSum + root.val);
    }
}