[LintCode] 中序遍历和后序遍历树构造二叉树 Construct Binary Tree from Inorder and Postorder Traversal

根据中序遍历和后序遍历树构造二叉树
注意事项
你可以假设树中不存在相同数值的节点
样例
给出树的中序遍历： [1,2,3] 和后序遍历： [1,3,2]
返回如下的树：
2
/ \
1 3

Given inorder and postorder traversal of a tree, construct the binary tree.
Notice
You may assume that duplicates do not exist in the tree.
Example
Given inorder [1,2,3] and postorder [1,3,2], return a tree:
2
/ \
1 3

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */


public class Solution {
    /**
     *@param inorder : A list of integers that inorder traversal of a tree
     *@param postorder : A list of integers that postorder traversal of a tree
     *@return : Root of a tree
     */
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(null == inorder || null == postorder || inorder.length != postorder.length || inorder.length <= 0) return null;
        return createTree(inorder, 0, inorder.length-1, postorder, 0, postorder.length-1);

    }

    public TreeNode createTree(int[] inorder, int il, int ir, int[] postorder, int pl,int pr) {
        if(il > ir || pl > pr) return null;
        TreeNode root = new TreeNode(postorder[pr]);
        int inlo = searchArray(inorder, postorder[pr]);
        root.left = createTree(inorder, il, inlo-1, postorder, pl,pl+inlo-il-1);
        root.right = createTree(inorder, inlo+1, ir, postorder, pl+inlo-il, pr-1);
        return root;
    }

    public int searchArray(int[] nums, int key) {//key一定在nums内
        int i = 0;
        while(nums[i++] != key);
        return --i;
    }
}