剑指Offer之Java算法习题精讲二叉树与N叉树
时间:2022-10-11 12:11:46|栏目:JAVA代码|点击: 次
题目一
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 { StringBuffer sb = new StringBuffer(); List<String> list = new ArrayList<String>(); public List<String> binaryTreePaths(TreeNode root) { method(root); return list; } public void method(TreeNode root){ if(root==null) return; int t = sb.length(); sb.append(root.val); if(root.left==null&&root.right==null){ list.add(sb.toString()); } sb.append("->"); method(root.left); method(root.right); sb.delete(t, sb.length()); } }
题目二
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 { int ans = 0; public int sumOfLeftLeaves(TreeNode root) { method(root,false); return ans; } public void method(TreeNode root,boolean flag){ if(root==null) return; if(root.left==null&&root.right==null&&flag){ ans+=root.val; return; } method(root.left,true); method(root.right,false); } }
题目三
解法
/* // Definition for a Node. class Node { public int val; public List<Node> children; public Node() {} public Node(int _val) { val = _val; } public Node(int _val, List<Node> _children) { val = _val; children = _children; } }; */ class Solution { public int maxDepth(Node root) { if(root==null){ return 0; } int maxChildDepth = 0; for(int i = 0;i<root.children.size();i++){ int childDepth = maxDepth(root.children.get(i)); maxChildDepth = Math.max(maxChildDepth, childDepth); } return maxChildDepth+1; } }