leetcode — construct-binary-tree-from-preorder-and-inorder-traversal

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
 *
 * Source : https://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
 *
 *
 * Given preorder and inorder traversal of a tree, construct the binary tree.
 *
 * Note:
 * You may assume that duplicates do not exist in the tree.
 */
public class ConstructFromPreorderAndInorder {


    /**
     * 使用前序和中序遍历结果来恢复一颗二叉树
     * preorder:root/left/right
     * inorder:left/root/right
     *
     * 根据preorder找到root,然后根据inorder找到left,right
     *
     * @param preorderArr
     * @param inorderArr
     * @return
     */
    public TreeNode build (char[] preorderArr, char[] inorderArr) {
        return buildByRecursion(preorderArr, 0, preorderArr.length-1, inorderArr, 0, inorderArr.length-1);
    }

    public TreeNode buildByRecursion (char[] preorerArr, int preStart, int preEnd, char[] inorderArr, int inStart, int inEnd) {
        if(preStart > preEnd || inStart > inEnd) {
            return null;
        }

        TreeNode root = new TreeNode(preorerArr[preStart] - '0');
        int rootIndex = -1;
        for (int i = inStart; i <= inEnd; i++) {
            if (preorerArr[preStart] == inorderArr[i]) {
                rootIndex = i;
                break;
            }
        }

        if (rootIndex < 0) {
            return null;
        }
        int leftTreeSize = rootIndex - inStart;
        int rightTreeSize = inEnd - rootIndex;
        root.leftChild = buildByRecursion(preorerArr, preStart+1, preStart + leftTreeSize,
                inorderArr, inStart, rootIndex-1);
        root.rightChild = buildByRecursion(preorerArr,preEnd-rightTreeSize+1, preEnd,
                inorderArr, rootIndex+1, inEnd );
        return root;

    }

    /**
     * 使用广度优先遍历将数转化为数组
     *
     * @param root
     * @param chs
     */
    public void binarySearchTreeToArray (TreeNode root, List<Character> chs) {
        if (root == null) {
            chs.add('#');
            return;
        }
        List<TreeNode> list = new ArrayList<TreeNode>();
        int head = 0;
        int tail = 0;
        list.add(root);
        chs.add((char) (root.value + '0'));
        tail ++;
        TreeNode temp = null;

        while (head < tail) {
            temp = list.get(head);
            if (temp.leftChild != null) {
                list.add(temp.leftChild);
                chs.add((char) (temp.leftChild.value + '0'));
                tail ++;
            } else {
                chs.add('#');
            }
            if (temp.rightChild != null) {
                list.add(temp.rightChild);
                chs.add((char)(temp.rightChild.value + '0'));
                tail ++;
            } else {
                chs.add('#');
            }
            head ++;
        }
        //去除最后不必要的
        for (int i = chs.size()-1; i > 0; i--) {
            if (chs.get(i) != '#') {
                break;
            }
            chs.remove(i);
        }
    }

    private class TreeNode {
        TreeNode leftChild;
        TreeNode rightChild;
        int value;

        public TreeNode(int value) {
            this.value = value;
        }

        public TreeNode() {

        }
    }

    public static void main(String[] args) {
        /*
         *     3
         *    / \
         *   9  2
         *     /  \
         *    1   7
         */

        char[] preorderArr = new char[]{'3','9','2','1','7'};
        char[] inorderArr = new char[] {'9','3','1','2','7'};

        ConstructFromPreorderAndInorder constructFromPreorderAndInorder = new ConstructFromPreorderAndInorder();

        TreeNode root = constructFromPreorderAndInorder.build(preorderArr, inorderArr);
        List<Character> chs = new ArrayList<Character>();
        constructFromPreorderAndInorder.binarySearchTreeToArray(root, chs);
        System.out.println(Arrays.toString(chs.toArray(new Character[chs.size()])));




    }

}
posted @ 2017-11-09 08:18  lacker  阅读(215)  评论(0编辑  收藏  举报