Leetcode 题目解析之 Construct Binary Tree from Preorder and Inorder T

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

Note:

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

题目要求：根据前序遍历和中序遍历序列，构建二叉树。

• 前序遍历p：1 2 4 3 5 6
• 中序遍历i：4 2 1 5 3 6

    1
   / \
  2   3
 /   / \
4   5   6

前序遍历p0是二叉树的根结点，1是i2，则：

• p1…2是1的左子树的前序遍历，i0…1是1的左子树的中序遍历
• p3…5是1的右子树的前序遍历，i3…5是1的右子树的中序遍历
• 递归

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;
    }