剑指Offer之Java算法习题精讲二叉树与斐波那契函数

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(inde武汉有哪些好玩的地方x==k){            ans = root.val;            return;        }        method(root.right,k);    }}

题目三

解法

/** * definition for长征意义 a binary tree node. * public class treenode { *     int val; *     treenode left; *     treenode right; *     treenode() {} *     treenode(金琥int val) { this.val = val; } *     treenode(int val, treenode left, treenode right) { *         this.val = val; *         this.left = left; *         this.right = right; *     } * } */c曹阳飞宇lass solution {    public int mindept夏奶奶h(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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