剑指Offer之Java算法习题精讲二叉树与N叉树
作者：明天一定.
跟着思路走，之后从简单题入手，反复去看，做过之后可能会忘记，之后再做一次，记不住就反复做，反复寻求思路和规律，慢慢积累就会发现质的变化
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;
}
}
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