Construct Binary Tree from Preorder and Inorder Traversal （medium）

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【题目】

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

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

【题意】

根据先序遍历和中序遍历创建二叉树即从前序遍历和中序遍历的结果重建一颗二叉树。

【分析】

There is an example.

        _______7______       /              \    __10__          ___2   /      \        /   4       3      _8            \    /             1  11

The preorder and inorder traversals for the binary tree above is:

preorder = {7,10,4,3,1,2,8,11}inorder = {4,10,3,1,7,11,8,2}

The first node in preorder alwasy the root of the tree. We can break the tree like:
1st round:
preorder: {7}, {10,4,3,1}, {2,8,11}
inorder: {4,10,3,1}, {7}, {11, 8,2}

        _______7______       /              \    {4,10,3,1}       {11,8,2}

Since we alreay find that {7} will be the root, and in "inorder" sert, all the data in the left of {7} will construct the left sub-tree. And the right part will construct a right sub-tree. We can the left and right part agin based on the preorder.

2nd round
left part right part
preorder: {10}, {4}, {3,1} {2}, {8,11}
inorder: {4}, {10}, {3,1} {11,8}, {2}

        _______7______       /              \    __10__          ___2   /      \        /   4      {3,1}   {11,8}

see that, {10} will be the root of left-sub-tree and {2} will be the root of right-sub-tree.

Same way to split {3,1} and {11,8}, yo will get the complete tree now.

        _______7______       /              \    __10__          ___2   /      \        /   4       3      _8            \    /             1  11

So, simulate this process from bottom to top with recursion as following code.

【实现】

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public TreeNode buildTree(int[] preorder, int[] inorder) {        return buildPI(preorder, inorder, 0, preorder.length-1, 0, inorder.length-1);    }<span style="white-space:pre"></span>public TreeNode buildPI(int[] preorder, int[] inorder, int p_s, int p_e, int i_s, int i_e){<span style="white-space:pre"></span>if(p_s>p_e)<span style="white-space:pre"></span>return null;<span style="white-space:pre"></span>int pivot = preorder[p_s];<span style="white-space:pre"></span>int i = i_s;<span style="white-space:pre"></span>for(;i<i_e;i++){<span style="white-space:pre"></span>if(inorder[i]==pivot)<span style="white-space:pre"></span>break;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>TreeNode node = new TreeNode(pivot);<span style="white-space:pre"></span>int lenLeft = i-i_s;<span style="white-space:pre"></span>node.left = buildPI(preorder, inorder, p_s+1, p_s+lenLeft, i_s, i-1);<span style="white-space:pre"></span>node.right = buildPI(preorder, inorder, p_s+lenLeft+1, p_e, i+1, i_e);<span style="white-space:pre"></span>return node;<span style="white-space:pre"></span>}

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