文章目录

• 力扣算法学习day13-1
• • 110-平衡二叉树
  • • 题目
    • 代码实现
  • 257-二叉树的所有路径
  • • 题目
    • 代码实现

力扣算法学习day13-1

110-平衡二叉树

题目

代码实现

/**
 * 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) {
        int height = height(root);
        if(height == -1){
            return false;
        } else{
            return true;
        }
    }
    // -1表示已近出现不符合平衡二叉树的存在。
    public int height(TreeNode node){
        if(node == null){// 返回高度为0
            return 0;
        }
        int left = height(node.left);
        int right = height(node.right);
        if(left == -1 || right == -1){
            return -1;
        }
        if(Math.abs(left - right) > 1){
            return -1;
        }
        int length = Math.max(left,right) + 1;

        return length;
    }
}

257-二叉树的所有路径

题目

代码实现

/**
 * 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> list = new ArrayList<>();
        String s = "";
        pathsList(root,s,list);
        return list;
    }
    public void pathsList(TreeNode node,String s,List<String> list){
        if("".equals(s)){
            s = s + node.val;
        }else{
            s = s + "->" + node.val;
        }
        
        if(node.left != null){
            pathsList(node.left,s,list);
        }
        if(node.right != null){
            pathsList(node.right,s,list);
        }

        if(node.left == null && node.right == null){
            list.add(s);
        }
    }

}