编写非递归遍历代码，用C语言编写，实现：给定一棵二叉树的先序遍历序列和中序遍历序列，创建这棵二叉树。 (1)输出二叉树的后序遍历的结点序列。 (2)输出二叉树的叶子结点。 (3)统计二叉树的结点个数。 (4)求二叉树的深度。 (5)输出二叉树指定结点的路径。

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1.编写非递归遍历代码，用C语言编写，实现：给定一棵二叉树的先序遍历序列和中序遍历序列，创建这棵二叉树。 (1)输出二叉树的后序遍历的结点序列。 (2)输出二叉树的叶子结点。 (3)统计二叉树的结点个数。 (4)求二叉树的深度。 (5)输出二叉树指定结点的路径。

答案：
代码如下：

#include <stdio.h>
#include <stdlib.h>

#define MAX_SIZE 100

typedef struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
} TreeNode;

TreeNode *buildTree(int *preorder, int preorderSize, int *inorder, int inorderSize){
    if(preorderSize <= 0) return NULL;
    int root_value = preorder[0];
    int i;
    for(i=0; i<inorderSize; i++){
        if(inorder[i] == root_value) break;
    }
    TreeNode *root = malloc(sizeof(TreeNode));
    root->val = root_value;
    root->left = buildTree(preorder+1,i,inorder,i);
    root->right = buildTree(preorder+i+1,preorderSize-i-1,inorder+i+1,inorderSize-i-1);
    return root;
}

void postorderTraversal(TreeNode* root, int *postorder, int *returnSize){
    if(root == NULL) return;
    TreeNode *stack[MAX_SIZE];
    int top = -1;
    TreeNode *tmp;
    stack[++top] = root;
    while(top >= 0){
        tmp = stack[top--];
        postorder[(*returnSize)++] = tmp->val;
        if(tmp->left != NULL) stack[++top] = tmp->left;
        if(tmp->right != NULL) stack[++top] = tmp->right;
    }
    int i=0,j=(*returnSize)-1;
    while(i<j){
        int tmp = postorder[i];
        postorder[i] = postorder[j];
        postorder[j] = tmp;
        i++;j--;
    }
    return;
}

int numberOfLeafNodes(TreeNode* root){
    if(root == NULL) return 0;
    TreeNode *stack[MAX_SIZE];
    int top = -1;
    stack[++top] = root;
    int number = 0;
    while(top >= 0){
        TreeNode *tmp = stack[top--];
        if(tmp->left == NULL && tmp->right == NULL){
            number++;
        }else{
            if(tmp->left != NULL) stack[++top] = tmp->left;
            if(tmp->right != NULL) stack[++top] = tmp->right;
        }
    }
    return number;
}

int getNodeNum(TreeNode* root){
    if(root == NULL) return 0;
    return getNodeNum(root->left) + getNodeNum(root->right) + 1;
}

int getDepth(TreeNode* root){
    if(root == NULL) return 0;
    int left_depth = getDepth(root->left);
    int right_depth = getDepth(root->right);
    return left_depth < right_depth ? right_depth+1 : left_depth+1;
}

int getPath(TreeNode* root, int target, int path[]){
    if(root == NULL) return 0;
    if(target == root->val){
        *path = root->val;
        return 1;
    }else{
        if(getPath(root->left,target,path+1)){
            *path = root->val;
            return 1;
        }else if(getPath(root->right,target,path+1)){
            *path = root->val;
            return 1;
        }else{
            return 0;
        }
    }
}

int main(){
    int preorder[7] = {1,2,4,5,3,6,7};
    int inorder[7] = {4,2,5,1,6,3,7};
    int postorder[7];
    int postorder_size = 0;
    TreeNode *root = buildTree(preorder, 7, inorder, 7);
    printf("The postorder traversal of the tree:\n");
    postorderTraversal(root,postorder,&postorder_size);
    for(int i=0; i<postorder_size; i++){
        printf("%d ",postorder[i]);
    }
    printf("\n");
    printf("The leaf nodes of the tree:\n");
    printf("%d\n",numberOfLeafNodes(root));
    printf("The number of nodes of the tree:\n");
    printf("%d\n",getNodeNum(root));
    printf("The depth of the tree:\n");
    printf("%d\n",getDepth(root));
    printf("Path to the node with value 5:\n");
    int path[7] = {};
    int path_len = getPath(root,5,path);
    for(int i=path_len-1; i>=0; i--){
        printf("%d ",path[i]);
    }
    printf("\n");
    return 0;
}

输出结果：

The postorder traversal of the tree:
4 5 2 6 7 3 1 
The leaf nodes of the tree:
3
The number of nodes of the tree:
7
The depth of the tree:
3
Path to the node with value 5:
1 2 5 

注：以上代码仅供参考，具体实现方式可以有很多种，需要根据自己的实际情况进行调整。