重建二叉树

题目：根据前序,中序重建二叉树

• 前序: 1 2 4 7 3 5 6 8
• 中序: 4 7 2 1 5 3 8 6

解析

1. 前序: 根左右 中序:左根右 后序:左右根
2. 由此可知：根节点为1，由中序可知左子树为472，右子树为5386
3. 由上面的分析可继续得出递归结果:例如只是左子树

4. java实现:

 /**
     * 已知 :
     * 前序: 1 2 4 7 3 5 6 8
     * 中序: 4 7 2 1 5 3 8 6
     */


    /**
     * Definition for binary tree
     */
    public class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode(int x) {
            val = x;
        }

        @Override
        public String toString() {
            return "TreeNode{" +
                    "val=" + val +
                    ", left=" + left +
                    ", right=" + right +
                    '}';
        }
    }


    public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
        TreeNode rootNode = new TreeNode(pre[0]);
        reConstruct(rootNode, pre, in);
        return rootNode;
    }

    public void reConstruct(TreeNode rootNode, int[] pre, int[] in) {


        //根节点
        int root = pre[0];
        //查找左节点 和 右节点
        //根节点所在位置
        int inRoot = 0;

        for (int i = 0; i < in.length; i++) {
            if (in[i] == root) {
                inRoot = i;
            }
        }

        //左子树
        if (inRoot > 0) {
            //中序
            int[] inVo = new int[inRoot];
            System.arraycopy(in, 0, inVo, 0, inRoot);

            //前序
            int[] preVo = new int[inRoot];
            System.arraycopy(pre, 1, preVo, 0, inRoot);
            rootNode.left = new TreeNode(preVo[0]);
            reConstruct(rootNode.left, preVo, inVo);
        }

        //右子树
        if (inRoot < in.length - 1) {
            //中序
            int[] inVo = new int[in.length - inRoot - 1];
            System.arraycopy(in, inRoot + 1, inVo, 0, in.length - inRoot - 1);

            //前序
            int[] preVo = new int[in.length - inRoot - 1];
            System.arraycopy(pre, inRoot + 1, preVo, 0, in.length - inRoot - 1);
            rootNode.right = new TreeNode(preVo[0]);
            reConstruct(rootNode.right, preVo, inVo);
        }

    }


    @Test
    public void test1() {
        System.out.println(reConstructBinaryTree(new int[]{1, 2, 4, 7, 3, 5, 6, 8}, new int[]{4, 7, 2, 1, 5, 3, 8, 6}));
    }