Question

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

Note:

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

Solution

参考：

Code

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* buildTree(vector
    
     & inorder, vector
     
      & postorder) {        if (inorder.size() == 0 || postorder.size() == 0)            return NULL;        return ConstructTree(inorder, postorder, 0, inorder.size() - 1, 0, postorder.size() - 1);    }        TreeNode* ConstructTree(vector
      
       & inorder, vector
       
        & postorder,             int in_start, int in_end, int post_start, int post_end) {        int rootValue = postorder[post_end];        TreeNode* root = new TreeNode(rootValue);                if (in_start == in_end) {            if (post_start == post_end && inorder[in_start] == postorder[post_start])                return root;        }                int rootIn = in_start;        while (rootIn <= in_end && inorder[rootIn] != rootValue)            rootIn++;                int leftPostLength = rootIn - in_start;        if (leftPostLength > 0) {            // 注意起点和终点的写法            root->left = ConstructTree(inorder, postorder, in_start, rootIn - 1, post_start, post_start + leftPostLength - 1);        }        if (leftPostLength < in_end - in_start) {            root->right = ConstructTree(inorder, postorder, rootIn + 1, in_end, post_start + leftPostLength, post_end - 1);        }                return root;    }};