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服务器之家 - 编程语言 - Java教程 - 剑指Offer之Java算法习题精讲二叉树与N叉树

剑指Offer之Java算法习题精讲二叉树与N叉树

2022-10-08 15:31明天一定. Java教程

跟着思路走,之后从简单题入手,反复去看,做过之后可能会忘记,之后再做一次,记不住就反复做,反复寻求思路和规律,慢慢积累就会发现质的变化

题目一

剑指Offer之Java算法习题精讲二叉树与N叉树

 解法

/**
* 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 {
  StringBuffer sb = new StringBuffer();
  List<String> list = new ArrayList<String>();
  public List<String> binaryTreePaths(TreeNode root) {
      method(root);
      return list;
  }
  public void method(TreeNode root){
      if(root==null) return;
      int t = sb.length();
      sb.append(root.val);
      if(root.left==null&&root.right==null){
          list.add(sb.toString());
      }
      sb.append("->");
      method(root.left);
      method(root.right);
      sb.delete(t, sb.length());
  }
}

 

题目二

剑指Offer之Java算法习题精讲二叉树与N叉树

 解法

/**
* 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 ans = 0;
  public int sumOfLeftLeaves(TreeNode root) {
      method(root,false);
      return ans;
  }
  public void method(TreeNode root,boolean flag){
      if(root==null) return;
      if(root.left==null&&root.right==null&&flag){
          ans+=root.val;
          return;
      }
      method(root.left,true);
      method(root.right,false);
  }
}

 

题目三

剑指Offer之Java算法习题精讲二叉树与N叉树

 解法

/*
// Definition for a Node.
class Node {
  public int val;
  public List<Node> children;
  public Node() {}
  public Node(int _val) {
      val = _val;
  }
  public Node(int _val, List<Node> _children) {
      val = _val;
      children = _children;
  }
};
*/

class Solution {
  public int maxDepth(Node root) {
      if(root==null){
          return 0;
      }
      int maxChildDepth = 0;
      for(int i = 0;i<root.children.size();i++){
          int childDepth = maxDepth(root.children.get(i));
          maxChildDepth = Math.max(maxChildDepth, childDepth);
      }
      return maxChildDepth+1;
  }
}

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原文链接:https://blog.csdn.net/wai_58934/article/details/123560260

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