題意:給出一個二叉樹的前序和中序遍歷,求出這個二叉樹
思路:首先枚舉中序遍歷找到根節點,之後再向左向右遞歸,並縮小遍歷範圍,若存在左節點左樹根就是start+1,右樹根位置可以根據當前根在中序和前序中的相對位置確定
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
return solve(preorder, inorder, 0, 0, preorder.length-1);
}
public TreeNode solve(int[] preorder, int[] inorder, int start, int l, int r) {
if (start >= preorder.length)
return null;
TreeNode root = new TreeNode(preorder[start]);
for (int i = l; i <= r; i++) {
if (inorder[i] == preorder[start]) {
if (i != l) {
root.left = solve(preorder, inorder, start+1, l, i-1);
}
if (i != r) {
root.right = solve(preorder, inorder, start+i-l+1, i+1, r);
}
break;
}
}
return root;
}
}