java

关注公众号 jb51net

关闭
首页 > 软件编程 > java > Java 二叉树构造

剑指Offer之Java算法习题精讲二叉树的构造和遍历

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

二叉树题——最大二叉树

根据给定的数组来构建最大二叉树

具体题目如下

 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 TreeNode constructMaximumBinaryTree(int[] nums) {
        return method(nums,0,nums.length-1);
    }
    public TreeNode method(int[] nums,int lo,int hi){
        if(lo>hi){
            return null;
        }
        int index = -1;
        int max = Integer.MIN_VALUE;
        for(int i = lo;i<=hi;i++){
            if(max<nums[i]){
                max = nums[i];
                index = i;
            }
        }
        TreeNode root = new TreeNode(max);
        root.left = method(nums,lo,index-1);
        root.right = method(nums,index+1,hi);
        return root;
    }
}

题目二

二叉树题——构造二叉树

根据给定的数组按照指定遍历条件构造二叉树并返回根节点

具体题目如下

 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 TreeNode buildTree(int[] preorder, int[] inorder) {
        return method(preorder,0,preorder.length-1,inorder,0,inorder.length-1);
    }
    public TreeNode method(int[] preorder, int preLeft,int preEnd , int[] inorder,int inLeft,int inEnd){
        if(preLeft>preEnd){
            return null;
        }
        int rootVal = preorder[preLeft];
        int index = -1;
        for(int i = inLeft;i<=inEnd;i++){
            if(rootVal == inorder[i]){
                index = i;
            }
        }
        TreeNode root = new TreeNode(rootVal);
        int leftSize = index - inLeft;
        root.left = method(preorder,preLeft+1,leftSize+preLeft,inorder,inLeft,index-1);
        root.right = method(preorder,leftSize+preLeft+1,preEnd,inorder,index+1,inEnd);
        return root;
    }
}

题目三

二叉树题——构造二叉树

根据给定的数组按照指定遍历条件构造二叉树并返回根节点

具体题目如下

 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 TreeNode buildTree(int[] inorder, int[] postorder) {
        return build(inorder,0,inorder.length-1,postorder,0,postorder.length-1);
    }
    TreeNode build(int[] inorder, int inStart, int inEnd,int[] postorder, int postStart, int postEnd) {
 
    if (inStart > inEnd) {
        return null;
    }
    // root 节点对应的值就是后序遍历数组的最后一个元素
    int rootVal = postorder[postEnd];
    // rootVal 在中序遍历数组中的索引
    int index = 0;
    for (int i = inStart; i <= inEnd; i++) {
        if (inorder[i] == rootVal) {
            index = i;
            break;
        }
    }
    // 左子树的节点个数
    int leftSize = index - inStart;
    TreeNode root = new TreeNode(rootVal);
    // 递归构造左右子树
    root.left = build(inorder, inStart, index - 1,postorder, postStart, postStart + leftSize - 1);
    root.right = build(inorder, index + 1, inEnd,postorder, postStart + leftSize, postEnd - 1);
    return root;
}
}

题目四

二叉树题——构造二叉树

根据给定的数组按照指定遍历条件构造二叉树并返回

具体题目如下

 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 TreeNode constructFromPrePost(int[] preorder, int[] postorder) {
        return method(preorder,0,preorder.length-1,postorder,0,postorder.length-1);
    }
    public TreeNode method(int[] preorder,int preStart, int preEnd, int[] postorder,int postStart,int postEnd){
        if(preStart>preEnd){
            return null;
        }
        if(preStart==preEnd){
            return new TreeNode(preorder[preStart]);
        }
        int rootVal = preorder[preStart];
        int leftRootVal = preorder[preStart + 1];
        int index = 0;
        for (int i = postStart; i < postEnd; i++) {
            if (postorder[i] == leftRootVal) {
                index = i;
                break;
            }
        }
        TreeNode root = new TreeNode(rootVal);
        int leftSize = index - postStart + 1;
        root.left = method(preorder, preStart + 1, preStart + leftSize,postorder, postStart, index);
        root.right = method(preorder, preStart + leftSize + 1, preEnd,postorder, index + 1, postEnd - 1);
        return root;
    }
}

到此这篇关于剑指Offer之Java算法习题精讲二叉树的构造和遍历的文章就介绍到这了,更多相关Java 二叉树构造内容请搜索脚本之家以前的文章或继续浏览下面的相关文章希望大家以后多多支持脚本之家!

您可能感兴趣的文章:
阅读全文