前言

选择一个简单的理念，矢志不渝地去执行（Take one simple idea and take it seriously

递归三部曲

确定递归函数的参数和返回值
确定终止条件
确定单层递归的逻辑

Leetcode 110.平衡二叉树

题目讲解

思路

深度指从根节点到该节点的最长简单路径边的条数。
高度指从该节点到叶子节点的最长简单路径边的条数。

这道题主要用到高度需要用后序算法
在这里插入图片描述

节点的左右子树的高度小于等于1

如果左右子树高度差大于一就返回 return -1;

/**
 * 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 boolean isBalanced(TreeNode root) {
      return getHeight(root)!=-1;   //传回的二叉树是二叉平衡树
    }
    public int getHeight(TreeNode root)
    {     
        if( root ==null)
        {
            return 0;
        }
        
        
         //后序遍历左右中
           int leftHeight =getHeight(root.left);
            if( leftHeight ==-1)
            {
                return -1;
            }
           int rightHeight =getHeight(root.right);
            if(rightHeight ==-1)
            {
                return-1;
            }  //用绝对值来避免 左 和右 前后顺序的差异
            if(Math.abs(leftHeight-rightHeight)>1)
            {
                return -1;     //如果节点左右子树的差大于一  则是不平衡二叉树
            }
            return Math.max( leftHeight,rightHeight)+1;
    }

}

Leetcode 257. 二叉树的所有路径

题目讲解

明天需要再看一遍

这道题涉及到了回溯

题解里讲到用了递归必然涉及到回溯
在这里插入图片描述
这里用的是前序遍历因为中左右从父结点到左右两个孩子在返回的时候就需要用到回溯减回去

/**
 * 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 List<String> binaryTreePaths(TreeNode root) {
        List<String> res =new ArrayList<>();
        if( res== null)
        {
            return res;
        }
        List<Integer> paths =new ArrayList<>();
        traveler(root,paths,res);
        return res;
    }
   public void traveler(TreeNode root,List<Integer>paths,List<String>res){
    //中
       paths.add(root.val);
    if( root.left ==null && root.right ==null)
    { 
       
       StringBuilder sb =new StringBuilder();
       for(int i=0;i<paths.size()-1;i++) 
       {
           sb.append(paths.get(i)).append("->");        
       }  
       sb.append(paths.get(paths.size()-1));             
       res.add(sb.toString());
       return;        
    }
    //左边
    if(root.left!=null)
    {
          traveler(root.left,paths,res);
          paths.remove(paths.size()-1);
    }
    //右边
    if(root.right!=null)
    {
         traveler(root.right,paths,res);
         paths.remove(paths.size()-1);
    }


}}

Leetcode 404.左叶子之和

题目讲解

nxp,java里居然没有value 只能用val

/**
 * 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 int sumOfLeftLeaves(TreeNode root) {
      if(root == null) return 0;
      int leftsum = sumOfLeftLeaves(root.left);
      int rightsum = sumOfLeftLeaves(root.right);
      int midsum =0;
      if(root.left!=null && root.left.left ==null&& root.left.right ==null)
      {
         midsum = root.left.val;
      }
      int sum = leftsum+rightsum+midsum;
      return sum;

    }
}