114. 二叉树展开为链表

题目描述

方法一；递归

//方法二：递归
class Solution {
    public void flatten(TreeNode root) {
        if (root == null) {
            return;
        }
        flatten(root.left);
        flatten(root.right);
        //到这里，按照结果来看，左右子树已经被捋成链表了，下面的只是逻辑，对于递归末端和上面任何一层都适用
        //写的时候就不需要再想末端情况了，直接想一种简单情况来搞通逻辑就行
        TreeNode tmp = root.right;
        root.right = root.left;//新的右子树为：把之前的左子树捋直了
        root.left = null;//新的左子树为null
        //找到新的右子树的末端
        TreeNode cur = root;
        while (cur.right != null) {
            cur = cur.right;
        }
        cur.right = tmp;//找到新的右子树的末端后，把捋直了的之前的右子树接上
    }
}

方法二：迭代

• 前序遍历整棵树，把每个节点添加到list中
• 从list中逐个取出节点添加到root.right

迭代版

class Solution {
    public void flatten(TreeNode root) {
        if (root == null) return; //特判
        Stack<TreeNode> stack = new Stack<>();
        List<TreeNode> list = new ArrayList<>();//存放前序遍历节点
        TreeNode node = root;
        //迭代，前序遍历
        while (node != null || !stack.isEmpty()) {
            if (node != null) {
                stack.push(node);
                list.add(node);
                node = node.left;
            } else {
                TreeNode father = stack.pop();
                node = father.right;
            }
        }
        //往root的right逐个添加节点
        root = list.get(0);
        TreeNode cur = root;
        for (int i = 1; i < list.size(); i++) {
            cur.right = list.get(i);
            cur.left = null;
            cur = cur.right;
        }
    }
}

递归版

class Solution {
    public void flatten(TreeNode root) {
        if(root == null) {
            return;//妈了个巴子，谁他妈没事干输入[]
        }
        List<TreeNode> list = new ArrayList<>();
        dfs(root, list);
        root = list.get(0);
        for (int i = 1; i < list.size(); i++) {
            root.right = list.get(i);
            root.left = null;//这个题需要将left置null
            root = root.right;
        }
    }
    public void dfs(TreeNode node, List<TreeNode> list) {
        if (node == null) {
            return;
        }
        list.add(node);
        dfs(node.left, list);
        dfs(node.right, list);
    }
}