重建二叉树-Java

2018-07-31  本文已影响8人  myserendipit
package algorithm;

import java.util.List;

public class RebuildBinaryTree {

    /**
     * 输入两个数组,分别表示的是前序遍历和中序遍历的结果
     *
     * @param preOrder 前序遍历 {1,2,4,7,3,5,6,8}
     * @param inOrder  中序遍历 {4,7,2,1,5,3,8,6}
     * @return 结果二叉树的根节点
     */
    public static TreeNode rebuild(int[] preOrder, int[] inOrder) {
        if (preOrder == null || inOrder == null ||
                preOrder.length != inOrder.length || inOrder.length <= 0) {
            return null;
        }


        int len = preOrder.length;
        TreeNode node = rebuildRecursive(preOrder, 0, inOrder, 0, len);
        return node;
    }

    public static TreeNode rebuildRecursive(int[] preOrder, int preStart,
                                            int[] inOrder, int inStart,
                                            int length) {
        if (length <= 0) return null;
        TreeNode<Integer> head = new TreeNode<>(preOrder[preStart]);
        int inOrderRootIndex = -1;
        for (int i = inStart; i < inStart + length; i++) {
            if (head.value == inOrder[i]) {
                // 找到了根节点的值(该根节点的意思也可以是子树的根节点)
                inOrderRootIndex = i;
                break;
            }
        }
        // 分为两堆数:左子树、右子树
        int leftLength = inOrderRootIndex - inStart;
        head.leftNode = rebuildRecursive(preOrder, preStart + 1, inOrder, inStart, leftLength);
        head.rightNode = rebuildRecursive(preOrder, preStart + leftLength + 1, inOrder, inOrderRootIndex + 1, length - leftLength - 1);

        return head;
    }
}
package algorithm;

public class TreeNode<T> {
    public T value;
    public TreeNode leftNode;
    public TreeNode rightNode;

    public TreeNode(T value) {
        this.value = value;
        leftNode = null;
        rightNode = null;
    }
}
上一篇下一篇

猜你喜欢

热点阅读