力扣labuladong——一刷day31

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二叉树解题的思维模式分两类： 1、是否可以通过遍历一遍二叉树得到答案？如果可以，用一个 traverse 函数配合外部变量来实现，这叫「遍历」的思维模式。 2、是否可以定义一个递归函数，通过子问题（子树）的答案推导出原问题的答案？如果可以，写出这个递归函数的定义，并充分利用这个函数的返回值，这叫「分解问题」的思维模式。 无论使用哪种思维模式，你都需要思考： 如果单独抽出一个二叉树节点，它需要做什么事情？需要在什么时候（前/中/后序位置）做？其他的节点不用你操心，递归函数会帮你在所有节点上执行相同的操作。

前言

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一、力扣226. 翻转二叉树

遍历思想

/**
 * 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 TreeNode invertTree(TreeNode root) {
        treaverse(root);
        return root;
    }
    public void treaverse(TreeNode root){
        if(root == null){
            return;
        }
        TreeNode l = root.left;
        root.left = root.right;
        root.right = l;
        treaverse(root.left);
        treaverse(root.right);
    }
}

分解思想

/**
 * 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 TreeNode invertTree(TreeNode root) {
        return fun(root);
    }
    public TreeNode fun(TreeNode root){
        if(root == null){
            return null;
        }
        TreeNode lchild = fun(root.left);
        TreeNode rchild = fun(root.right);
        root.left = rchild;
        root.right = lchild;
        return root;
    }
}

二、力扣116. 填充每个节点的下一个右侧节点指针

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}
    
    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _left, Node _right, Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/

class Solution {
    public Node connect(Node root) {
        if(root == null){
            return root;
        }
        if(root.left != null && root.right != null){
            fun(root.left, root.right);
        }
        return root;
    }
    public void fun(Node node1, Node node2){
        if(node1 == null || node2 == null){
            return ;
        }
        node1.next = node2;
        fun(node1.left, node1.right);
        fun(node2.left,node2.right);
        fun(node1.right,node2.left);
    }
}

三、力扣114. 二叉树展开为链表

/**
 * 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 void flatten(TreeNode root) {
        fun(root);
    }
    public TreeNode fun(TreeNode root){
        if(root == null){
            return null;
        }
        TreeNode r1 = fun(root.left);
        TreeNode r2 = fun(root.right);
        if(r1 != null && r2 != null){
            r1.right = root.right;
            root.right = root.left;
            root.left = null;
            return r2;
        }
        if(r1 == null && r2 != null){
            return r2;
        }
        if(r2 == null && r1 != null){
            root.right = root.left;
            root.left = null;
            return r1;
        }
        return root;
    }
}

第二种解法

/**
 * 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 {
    public void flatten(TreeNode root) {
        if(root == null){
            return;
        }
        TreeNode r1 = root.left;
        TreeNode r2 = root.right;
        flatten(root.left);
        flatten(root.right);
        root.left = null;
        root.right = r1;
        TreeNode p = root;
        while(p.right != null){
            p = p.right;
        }
        p.right = r2;
    }
}