从中序与后序遍历序列构造二叉树(Java)

从中序与后序遍历序列构造二叉树(Java)

大体思路：从后序数组中找到最后一个元素的值，即为当前节点并进行节点的创建，并在中序数组中找到该值所在索引（使用Map）。接着开始递归，后序数组倒着向前（所以递归应该先从右子树开始），当右子树遍历完（即不存在遍历区间时left>right）return。

问题分析：

1. 返回值：返回最终头节点；
2. 参数：由于需要遍历区间，所以定义两个参数left、right，至于map, postorder作为全局变量使用。

/**
 * 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 {
    Map<Integer, Integer> map = new HashMap<>();
    int[] postorder;
    int postindex;
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        this.postorder = postorder;
        postindex = postorder.length - 1;
        for(int i = 0; i < inorder.length; i++){
            map.put(inorder[i], i); // 通过值找索引（而且题目中表示值不会相同）
        }
        return treeBuild(0, postindex);
    }
    public TreeNode treeBuild(int left, int right //注意这个是inorder的左右区间 ){
        if(left > right){
            return null;
        }
        int rootval = postorder[postindex--];
        TreeNode root = new TreeNode(rootval); //创建节点
        int mid = map.get(rootval); //中序的切割点
        root.right = treeBuild(mid + 1, right);
        root.left = treeBuild(left, mid - 1);
        return root;
    }
}

从前序与中序遍历序列构造二叉树(Java)

思路分析：整体思路类似上一题，只不过这次先序遍历数组中的顺序是中左右，所以根节点是在最前边，同时递归时注意先左子树后右子树

/**
 * 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 {
    Map<Integer, Integer> map = new HashMap<>();
    int[] preorder;
    int i = 0; // 记录先序数组中的中间节点的索引
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        this.preorder = preorder;
        for(int i = 0; i < inorder.length; i++){
            map.put(inorder[i], i);
        }
        return fucTree(0, preorder.length - 1);

    }
    public TreeNode fucTree(int left, int right){
        if(left > right){
            return null;
        }
        int rootval = preorder[i++];
        TreeNode root = new TreeNode(rootval);
        int mid = map.get(rootval);
        root.left = fucTree(left, mid - 1);
        root.right = fucTree(mid + 1, right);
        return root;
    }
}