力扣labuladong——一刷day47

原创已于 2023-11-22 18:41:31 修改 · 460 阅读

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#leetcode #算法 #java #数据结构

于 2023-11-22 18:39:19 首次发布

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文章介绍了四个与二叉树相关的LeetCode题目，涉及堂兄弟节点判断、祖父节点值为偶数的节点计数、好节点统计以及寻找独生节点，主要展示了遍历和递归在解决这类问题中的应用。

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文章目录

前言
一、力扣993. 二叉树的堂兄弟节点
二、力扣1315. 祖父节点值为偶数的节点和
三、力扣1448. 统计二叉树中好节点的数目
四、力扣1469. 寻找所有的独生节点

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「遍历」的思维。

一、力扣993. 二叉树的堂兄弟节点

/**
 * 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 {
    int depthX = -1, depthY = -1;
    boolean flag = true;
    public boolean isCousins(TreeNode root, int x, int y) {
        fun(root,1,x,y);
        if(flag == false){
            return false;
        }
        if(depthX == depthY){
            return true;
        }
        return false;
    }
    public void fun(TreeNode root, int depth, int x, int y){
        if(root == null){
            return ;
        }
        if(root.val == x){
            depthX = depth;
        }
        if(root.val == y){
            depthY = depth;
        }
        if(root.left != null && root.right != null){
            if((root.left.val == x || root.left.val == y) && (root.right.val == x || root.right.val == y)){
                flag = false;
            }
        }
        fun(root.left,depth+1,x,y);
        fun(root.right,depth+1,x,y);
    }
}

二、力扣1315. 祖父节点值为偶数的节点和

/**
 * 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 sum = 0;
    public int sumEvenGrandparent(TreeNode root) {
        fun(root, null, null);
        return sum;
    }
    public void fun(TreeNode root, TreeNode parent, TreeNode grandParent){
        if(root == null){
            return;
        }
        if(grandParent != null && grandParent.val % 2 == 0){
            sum += root.val;
        }
        fun(root.left, root, parent);
        fun(root.right, root, parent);
    }
}

三、力扣1448. 统计二叉树中好节点的数目

/**
 * 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;
    public int goodNodes(TreeNode root) {
        fun(root,Integer.MIN_VALUE);
        return count;
    }
    public void fun(TreeNode root, int preMax){
        if(root == null){
            return;
        }
        if(root.val >= preMax){
            preMax = root.val;
            count ++;
        }
        fun(root.left, preMax);
        fun(root.right, preMax);
    }
}

四、力扣1469. 寻找所有的独生节点

/**
 * 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<Integer> res = new ArrayList<>();
    public List<Integer> getLonelyNodes(TreeNode root) {
        fun(root);
        return res;
    }
    public void fun(TreeNode root){
        if(root == null){
            return;
        }
        if(root.left == null && root.right != null){
            res.add(root.right.val);
        }
        if(root.left != null && root.right == null){
            res.add(root.left.val);
        }
        fun(root.left);
        fun(root.right);
    }
}