94. 二叉树的中序遍历

class Solution {
    List<Integer> res = new ArrayList<>();
    public List<Integer> inorderTraversal(TreeNode root) {
        if(root==null) return res;
        inorderTraversal(root.left);
        res.add(root.val);
        inorderTraversal(root.right);
        return res;
    }
}

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        TreeNode curr = root;
        while(curr != null || !stack.isEmpty()){
            if(curr!=null){
                stack.push(curr);
                curr = curr.left;
            }else{
                curr = stack.pop();
                res.add(curr.val);
                curr = curr.right;
            }
        }
        return res;
    }
}

114. 二叉树展开为链表

class Solution {
    public void flatten(TreeNode root) {
        while(root!=null){
            if(root.left==null){
                root = root.right;
            }else{
                TreeNode pre = root.left;
                while(pre.right!=null){
                    pre = pre.right;
                }
                pre.right = root.right;

                root.right = root.left;
                root.left = null;

                root = root.right;
            }
        }
    }
}

class Solution {
    public void flatten(TreeNode root) {
        if(root == null){
            return ;
        }
        //将根节点的左子树变成链表
        flatten(root.left);
        //将根节点的右子树变成链表
        flatten(root.right);
        TreeNode temp = root.right;
        //把树的右边换成左边的链表
        root.right = root.left;
        //记得要将左边置空
        root.left = null;
        //找到树的最右边的节点
        while(root.right != null) root = root.right;
        //把右边的链表接到刚才树的最右边的节点
        root.right = temp;
    }
}

1. 两数之和

class Solution {
    public int[] twoSum(int[] nums, int target) {
        for(int i = 0;i < nums.length;i++){
            int s  = target - nums[i];
            for(int j = i+1;j < nums.length;j++){
                if(s==nums[j]){
                    return new int[]{i,j};
                }
            }
        }
        return new int[]{};
    }
}

class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer,Integer> map = new HashMap<>();
        int[] res = new int[2];
        for(int i = 0;i < nums.length;i++){
            if(map.containsKey(target - nums[i])){
                res[0] = map.get(target - nums[i]);
                res[1] = i;
                return res;
            }
            map.put(nums[i],i);
        }
        return res;
    }
}

你知道的越多，你不知道的越多。

标签：---,right,curr,2020.9,int,res,力扣,null,root
来源： https://blog.csdn.net/qq_40722827/article/details/108591133