[ 二叉树 ] 中序和后序遍历确定二叉树

106. 从中序与后序遍历序列构造二叉树 - 力扣(LeetCode) (leetcode-cn.com)

中序和后序遍历确定二叉树

递归

  • 明确:分治
  • 左闭右开切割
class Solution {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        // 构造返回值
        TreeNode* root = new TreeNode();

        // 空组返回
        if (postorder.empty()) return nullptr;

        // 寻找中序切割点, 也同样为根节点
        int cutPointValue = postorder[postorder.size() - 1];
        root->val = cutPointValue;

        // 如果只剩一个根结点直接返回即可
        if (postorder.size() == 1) return root;

        // 否则切割
        int inCutPoint = 0;
        for (; inCutPoint < inorder.size(); ++inCutPoint)
            if (cutPointValue == inorder[inCutPoint]) break;

        // 左闭右开原则
        vector<int> inLeft(inorder.begin(), inorder.begin() + inCutPoint);
        vector<int> inRight(inorder.begin() + inCutPoint + 1, inorder.end());

        int postCutPoint = inLeft.size();
        vector<int> postLeft(postorder.begin(), postorder.begin() + postCutPoint);
        vector<int> postRight(postorder.begin() + postCutPoint, postorder.end() - 1);

        root->left = buildTree(inLeft, postLeft);
        root->right = buildTree(inRight, postRight);

        return root;
    }
};
  • 空间优化 利用下标代替切割
  • 如果每次切割都创造新向量则会产生过多的空间开销
class Solution {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        return traversal(inorder, 0, inorder.size(), postorder, 0, postorder.size());
    }

    TreeNode* traversal(vector<int> &inorder, int inBegin, int inEnd,
        vector<int> &postOrder, int postBegin, int postEnd) {
        if (postBegin == postEnd) return nullptr;

        int rootValue = postOrder[postEnd - 1];
        TreeNode* root = new TreeNode(rootValue);

        if (postBegin == postEnd - 1) return root;

        int inCut = 0;
        for (; inCut < inEnd; ++inCut) {
            if (inorder[inCut] == rootValue) break;
        }
		
        // 传递下标代替切割
        int leftInBegin = inBegin;
        int leftInEnd = inCut;
        int rightInBegin = inCut + 1;
        int rightInEnd = inEnd;
        int leftPostBegin = postBegin;
        int leftPostEnd = postBegin + leftInEnd - leftInBegin;
        int rightPostBegin = leftPostEnd;
        int rightPostEnd = postEnd - 1;

        root->left = traversal(inorder, leftInBegin, leftInEnd, postOrder, leftPostBegin, leftPostEnd);
        root->right = traversal(inorder, rightInBegin, rightInEnd, postOrder, rightPostBegin, rightPostEnd);
        return root;
    }
};

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