C++实现LeetCode(106.由中序和后序遍历建立二叉树)

[LeetCode] 106. Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树

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

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

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

    3
/ \
9  20
/  \
15   7

这道题要求从中序和后序遍历的结果来重建原二叉树,我们知道中序的遍历顺序是左-根-右,后序的顺序是左-右-根,对于这种树的重建一般都是采用递归来做,可参见博主之前的一篇博客 Convert Sorted Array to Binary Search Tree。针对这道题,由于后序的顺序的最后一个肯定是根,所以原二叉树的根结点可以知道,题目中给了一个很关键的条件就是树中没有相同元素,有了这个条件就可以在中序遍历中也定位出根节点的位置,并以根节点的位置将中序遍历拆分为左右两个部分,分别对其递归调用原函数。代码如下:

class Solution {
public:
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        return buildTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);
    }
    TreeNode *buildTree(vector<int> &inorder, int iLeft, int iRight, vector<int> &postorder, int pLeft, int pRight) {
        if (iLeft > iRight || pLeft > pRight) return NULL;
        TreeNode *cur = new TreeNode(postorder[pRight]);
        int i = 0;
        for (i = iLeft; i < inorder.size(); ++i) {
            if (inorder[i] == cur->val) break;
        }
        cur->left = buildTree(inorder, iLeft, i - 1, postorder, pLeft, pLeft + i - iLeft - 1);
        cur->right = buildTree(inorder, i + 1, iRight, postorder, pLeft + i - iLeft, pRight - 1);
        return cur;
    }
};

上述代码中需要小心的地方就是递归是 postorder 的左右 index 很容易写错,比如 pLeft + i - iLeft - 1, 这个又长又不好记,首先我们要记住 i - iLeft 是计算 inorder 中根节点位置和左边起始点的距离,然后再加上 postorder 左边起始点然后再减1。我们可以这样分析,如果根结点就是左边起始点的话,那么拆分的话左边序列应该为空集,此时 i - iLeft 为0, pLeft + 0 - 1 < pLeft, 那么再递归调用时就会返回 NULL, 成立。如果根节点是左边起始点紧跟的一个,那么 i - iLeft 为1, pLeft + 1 - 1 = pLeft,再递归调用时还会生成一个节点,就是 pLeft 位置上的节点,为原二叉树的一个叶节点。

下面来看一个例子, 某一二叉树的中序和后序遍历分别为:

Inorder:    11  4  5  13  8  9

Postorder:  11  4  13  9  8  5  

11  4  5  13  8  9      =>          5

11  4  13  9  8  5                /  \

11  4     13   8  9      =>         5

11  4     13  9  8                  /  \

                             4   8

11       13    9        =>         5

11       13    9                    /  \

                             4   8

                            /    /     \

                           11    13    9

Github 同步地址:

https://github.com/grandyang/leetcode/issues/106

类似题目:

Construct Binary Tree from Preorder and Postorder Traversal

Construct Binary Tree from Preorder and Inorder Traversal

参考资料:

https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/discuss/758462/C%2B%2B-Detail-Explain-or-Diagram

https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/discuss/34803/Sharing-my-straightforward-recursive-solution

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