[Leetcode series] constructing a binary tree (iterative solution) from the central and post-order traversal sequences)

For a given medium-order traversal of inorder and post-order traversal of postorder, construct a binary tree.

Algorithm idea: Set post-order traversal to Po, and Middle-order traversal to Io.

 

• First, the last node of the Po is taken as the root node, and the node is added to the STN stack;
• Then compare the last node of Io with the top node of the STN Stack:
  • If the node is different, add the node to the right of the top node of the stack to the STN stack, and delete the node from the Po;
    • At this time, the stack stores all the right root nodes that have not yet processed the left subtree.
    • When a difference occurs, the depth of the right subtree increases by 1, and the stack depth represents the depth of the current right subtree.
  • If they are the same, first cache the stack top node and delete the IO and stack top elements respectively:
    • If it is still the same (indicating that there is no left subtree in this layer), step 2 is returned;
    • If they are different (the left subtree is included), add the last node of the Po to the left of the P cache node, and add the left subtree to the stack and delete the node from the Po, return to step 2;

Code:

 1 /** 2  * Definition for binary tree 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12     TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {13         if(inorder.size() == 0)return NULL;14         TreeNode *p;15         TreeNode *root;16         stack<TreeNode *> stn;17         18         root = new TreeNode(postorder.back()); 19         stn.push(root); 20         postorder.pop_back(); 21         22         while(true)23         {24             if(inorder.back() == stn.top()->val) 25             {26                 p = stn.top();27                 stn.pop(); 28                 inorder.pop_back(); 29                 if(inorder.size() == 0) break;30                 if(stn.size() && inorder.back() == stn.top()->val)31                     continue;32                 p->left = new TreeNode(postorder.back()); 33                 postorder.pop_back();34                 stn.push(p->left);35             }36             else 37             {38                 p = new TreeNode(postorder.back());39                 postorder.pop_back();40                 stn.top()->right = p; 41                 stn.push(p); 42             }43         }44         return root;45     }46 };

 

[Leetcode series] constructing a binary tree (iterative solution) from the central and post-order traversal sequences)