43 lines
1.1 KiB
C
43 lines
1.1 KiB
C
// 二叉树中序遍历(给出每个结点的左右孩子编号,根为1)
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// 输入:n,随后 n 行,每行两个整数 l、r(0 表示空)。
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// 输出:中序遍历序列,每个数字后均输出一个空格。
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#include <stdio.h>
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#include <stdlib.h>
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int main(void) {
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int n;
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if (scanf("%d", &n) != 1) return 0;
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int *L = (int *)malloc((n + 1) * sizeof(int));
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int *R = (int *)malloc((n + 1) * sizeof(int));
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if (!L || !R) return 0;
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for (int i = 1; i <= n; ++i) {
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int l, r;
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if (scanf("%d %d", &l, &r) != 2) {
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free(L); free(R);
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return 0;
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}
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L[i] = l; R[i] = r;
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}
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// 迭代式中序遍历:一路向左压栈,弹栈访问,再转向右子树
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int *stack = (int *)malloc((n + 1) * sizeof(int));
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int top = 0;
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int cur = 1; // 根节点编号为 1
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while (cur || top) {
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while (cur) { // 沿左链入栈
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stack[top++] = cur;
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cur = L[cur];
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}
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int u = stack[--top];
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printf("%d ", u);
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cur = R[u];
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}
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free(stack);
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free(L); free(R);
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return 0;
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} |