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// 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: | |
// Construct Binary Tree from Preorder and Inorder Traversal | |
// 由先序和中序重建二叉树 | |
TreeNode* buildTreeInPre(vector<int>& preorder, vector<int>& inorder) { | |
if (preorder.empty()) | |
return NULL; | |
stack<TreeNode *> st; | |
int pt = 0, it = 0; | |
bool flag = false; | |
TreeNode *pnode = new TreeNode(preorder[pt++]), *root = pnode; | |
st.push(pnode); | |
while (pt < preorder.size() || it < inorder.size()) { | |
if (!st.empty() && st.top()->val == inorder[it]) { | |
pnode = st.top(); st.pop(); | |
it ++; | |
flag = true; | |
} | |
else { | |
TreeNode* cur = new TreeNode(preorder[pt++]); | |
if (flag) { | |
pnode->right = cur; | |
flag = false; | |
} | |
else { | |
st.top()->left = cur; | |
} | |
st.push(cur); | |
} | |
} | |
return root; | |
} | |
// Construct Binary Tree from Inorder and Postorder Traversal | |
// 由后序和中序重建二叉树 | |
TreeNode* buildTreeInPost(vector<int>& inorder, vector<int>& postorder) { | |
if (inorder.empty()) | |
return NULL; | |
stack<TreeNode *> st; | |
int pt = postorder.size() - 1, it = pt; | |
TreeNode *pnode = new TreeNode(postorder[pt--]), *root = pnode; | |
st.push(pnode); | |
bool flag = false; | |
while (pt >= 0 || it >= 0) { | |
if (!st.empty() && st.top()->val == inorder[it]) { | |
pnode = st.top(); st.pop(); | |
flag = true; | |
it --; | |
} | |
else { | |
TreeNode* cur = new TreeNode(postorder[pt--]); | |
if(flag) { | |
pnode->left = cur; | |
flag = false; | |
} | |
else { | |
st.top()->right = cur; | |
} | |
st.push(cur); | |
} | |
} | |
return root; | |
} | |
}; |
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