题目描述

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

代码

    int p = 0;
    int[] preorder;
    int[] inorder;

    public TreeNode buildTree(int[] preorder, int[] inorder) {

        this.preorder = preorder;
        this.inorder = inorder;

        return buildTree(0, preorder.length);
    }

    TreeNode buildTree(int start, int end) {

        if (start >= end) {
            return null;
        }

        TreeNode root = new TreeNode(preorder[p]);

        int i;
        for (i = start; i < end && preorder[p] != inorder[i]; i++)
            ;

        p++;
        root.left = buildTree(start, i);
        root.right = buildTree(i + 1, end);

        return root;
    }