重建二叉树

106. Construct Binary Tree from Inorder and Postorder Traversal
105. Construct Binary Tree from Preorder and Inorder Traversal
由前序遍历+中序遍历（或中序遍历+后序遍历）重建二叉树。

• 以前序+中序为例，前序为根左右，因此第一个出现的为根节点，而中序则是左根右，根节点位于中间。因此可以在中序遍历中找根节点的位置，此位置左边为左子树，右边都为右子树。如此分别递归左子树与右子树的序列，即能重建二叉树。

/**
 - Definition for a binary tree node.
 - public class TreeNode {
 -     int val;
 -     TreeNode left;
 -     TreeNode right;
 -     TreeNode(int x) { val = x; }
 - }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
       return helper(preorder,0,preorder.length-1,inorder,0,inorder.length-1);
    }
    public TreeNode helper(int[] preorder,int startPre,int endPre,int[] inorder,int startIn,int endIn){
        if(startPre>endPre || startIn>endIn) return null;
        TreeNode root = new TreeNode(preorder[startPre]);
        for(int i=startIn;i<=endIn;i++){
            if(inorder[i] == preorder[startPre]){
                root.left = helper(preorder,startPre+1,startPre+i-startIn,inorder,startIn,i-1);
                root.right = helper(preorder,startPre+i-startIn+1,endPre,inorder,i+1,endIn);
                return root; 
            }
        }
        return null;
    }
}

• 中序遍历+后序遍历同理，后序为左右根，因此后序遍历从后往前的第一个即为根节点，在中序遍历中找此根节点，同理得到左子树与右子树，再分别递归：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        return helper(inorder,0,inorder.length-1,postorder,0,postorder.length-1);
    }
    public TreeNode helper(int[] inorder,int startIn,int endIn,int[] postorder,int startPost,int endPost){
        if(startIn>endIn || startPost>endPost) return null;
        TreeNode root = new TreeNode(postorder[endPost]);
        for(int i = startIn;i<=endIn;i++){
            if(inorder[i] == root.val){
                root.left = helper(inorder,startIn,i-1,postorder,startPost,startPost+i-startIn-1);
                root.right = helper(inorder,i+1,endIn,postorder,startPost+i-startIn,endPost-1);
                return root;
            }
        }
        return null;
    }
}