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思路#xff1a; 先序遍历#xff1a;根、左子树、右子树#xff1b;中序遍历#xff1a;左子树、根、右子树#xff1b;遍历先序遍历数组 prev#xff0c;使用一个辅助栈缓存“根节点”#xff1b;通过栈顶“根节点”与中序遍历数组 in 比较#xff0c;确认是否到…栈
思路 先序遍历根、左子树、右子树中序遍历左子树、根、右子树遍历先序遍历数组 prev使用一个辅助栈缓存“根节点”通过栈顶“根节点”与中序遍历数组 in 比较确认是否到了“最左”节点 如果没有到最左节点将 prev[idx] 节点挂到栈顶的左子树节点上并且入栈如果到了“最左”节点出栈直到不是“最左”节点将节点挂到栈顶的右子树节点上
/*** Definition for a binary tree node.* struct TreeNode {* int val;* TreeNode *left;* TreeNode *right;* TreeNode() : val(0), left(nullptr), right(nullptr) {}* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:TreeNode* buildTree(vectorint preorder, vectorint inorder) {if (!preorder.size()) {return nullptr;}TreeNode* root new TreeNode(preorder[0]);std::stackTreeNode* stk;stk.push(root);int in_idx 0;for (int idx 1; idx preorder.size(); idx) {int pre_val preorder[idx];TreeNode* node stk.top();if (node-val ! inorder[in_idx]) {node-left new TreeNode(pre_val);stk.push(node-left);} else {while (!stk.empty() (stk.top()-val inorder[in_idx])) {node stk.top();stk.pop();in_idx;}node-right new TreeNode(pre_val);stk.push(node-right);}}return root;}
};
