递归

To solve these two problems, I would like to first understand the idea of ​​recursion,There are three elements to solving recursive problems：

1. 方法参数和返回值

• 方法参数：There are two cases for method parameters
  • If it is related to the previous,Then you have to set it as a global variable
    例如刷题中的538题,就会出问题,As a result, the value passed in is not the latest,下面的543the same question.
  • If it doesn't matter before,That is to say, it only works on the current layer,即局部的,It is passed through the method body
• 返回值
  When you need to use the return value,设置返回值,The following two questions will be required,
  而上面提到的刷题中的538题,已经使用sumThe values ​​I need are logged,No return value operation is required.另外,Generally judged with return value,有false直接返回了.

2. 终止条件
   It's where you need to be clear,When to stop and when to end.
3. 函数体
   Operations on nodes and data

下面的题目,I will go through the above three steps to analyze

543. 二叉树的直径

543. 二叉树的直径
这道题翻译一下,Find the largest of the sum of the heights of the left and right subtrees of each node.
Why grab every node,Because I did it with hierarchical traversal at first,I thought it must start from the follow node,这个想法是错的,举个例子,The height of the root's right subtree is 1,The left subtree is two subtrees of the same height,高度都为10.
代码：

	//It is written outside because this is a global variable,Find the global largest.
	int ans;
    public  int diameterOfBinaryTree(TreeNode root) {
    
        depth(root);
        return ans;
    }
	//The return value is set because it is needed to find the height
    private  int depth(TreeNode root){
    
        //结束条件
        if(root == null){
    
            return 0;
        }
        //方法体
        // Recursively find the height of the left and right subtrees of the current node
        int leftDepth = depth(root.left);
        int rightDepth = depth(root.right);
        ans = Math.max(ans,leftDepth+rightDepth);
        //返回当前子树的高度
        return Math.max(leftDepth,rightDepth)+1;
    }

617. 合并二叉树

617. 合并二叉树
题目很好理解,思路,Pick up new ones

public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
    
        return build(root1, root2);
    }
    //先序遍历.
    //The return value is set because it is needed to build a tree
    public static TreeNode build(TreeNode root1, TreeNode root2){
    
		//终止条件
		//叶子结点,直接返回
        if(root1==null &&  root2==null) return null;
        //And wool scallops,可冲
        if(root1!=null && root2==null) return root1;
        if(root1==null && root2!=null) return root2;
        //The overlapping part of the left and right subtrees,需要叠加,
        TreeNode node = new TreeNode(root1.val + root2.val);
        //建树,Here you can see the return value that needs to be used.
        node.left=build(root1.left, root2.left);
        node.right=build(root1.right,root2.right);
        return node;
    }