105. 从前序与中序遍历序列构造二叉树

105. 从前序与中序遍历序列构造二叉树

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private Map<Integer, Integer> map;
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder == null || inorder == null || preorder.length != inorder.length)
            return null;
        int len = preorder.length;
        map = new HashMap<>();
        for (int i = 0; i < len; i++) 
            map.put(inorder[i], i);
        return build(preorder, inorder, 0, len - 1, 0, len - 1);
    }
    private TreeNode build(int[] preorder, int[] inorder, int preLeft, int preRight, int inLeft, int inRight) {
        if (preLeft > preRight)
            return null;
        // 先序遍历的第一个节点就是当前子树的根节点
        int rootVal = preorder[preLeft];
        TreeNode root = new TreeNode(rootVal);
        // 在中序遍历序列中找到根节点所在索引的位置
        int rootIndexInInOrder = map.get(rootVal);
        // 确定左右子树的节点的数目
        int leftNum = rootIndexInInOrder - inLeft;
        int rightNum = inRight - rootIndexInInOrder;
        root.left = build(preorder, inorder, preLeft + 1, preLeft + leftNum, inLeft, rootIndexInInOrder - 1);
        root.right = build(preorder, inorder, preLeft + leftNum + 1, preRight, rootIndexInInOrder + 1, rootIndexInInOrder + rightNum);
        return root;
    }
}