剑指Offer之Java算法习题精讲二叉树与斐波那契函数
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
class Solution {
public int fib(int n) {
int[] arr = new int[31];
arr[0] = 0;
arr[1] = 1;
for(int i = 2;i<=n;i++){
arr[i] = arr[i-2]+arr[i-1];
}
return arr[n];
}
}
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 index = 0;
int ans = 0;
public int kthSmallest(TreeNode root, int k) {
method(root,k);
return ans;
}
void method(TreeNode root, int k){
if(root==null) return;
method(root.left,k);
index++;
if(index==k){
ans = root.val;
return;
}
method(root.right,k);
}
}
题目三
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 {
public int minDepth(TreeNode root) {
if (root == null) {
return 0;
}
if (root.left == null && root.right == null) {
return 1;
}
int min_depth = Integer.MAX_VALUE;
if (root.left != null) {
min_depth = Math.min(minDepth(root.left), min_depth);
}
if (root.right != null) {
min_depth = Math.min(minDepth(root.right), min_depth);
}
return min_depth + 1;
}
}
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