演算法-二叉樹-1

阿新 • • 發佈：2021-07-14

先序遍歷非遞迴寫法

1. 先入棧彈出列印

2. 然後先壓入右節點再壓入左節點

3. 重複

 1 public void printPre(TreeNode node) {
 2         Stack<TreeNode> stack = new Stack<>();
 3         stack.push(node);
 4         while (!stack.isEmpty()) {
 5             TreeNode pop = stack.pop();
 6             System.out.println(pop.value);
 
 7             if (pop.right != null) {
 8                 stack.push(pop.right);
 9             }
10             if (pop.left != null) {
11                 stack.push(pop.left);
12             }
13 
14         }
15     }

中序遍歷非遞迴寫法

1. 先入棧所有的左子節點

2. 彈出節點輸出並壓入彈出節點的右節點的左子節點

 1 public void printMid(TreeNode node) {
 
 2         Stack<TreeNode> stack = new Stack<>();
 3         while (node != null || !stack.isEmpty()) {
 4             if (node != null) {
 5                 stack.push(node);
 6                 node = node.left;
 7             } else {
 8                 TreeNode pop = stack.pop();
 9                 System.out.println(pop.value);
 
10                 node = pop.right;
11             }
12         }
13 
14     }

後序遍歷非遞迴寫法

1. 兩個棧stack/result

2. 頭節點陷入stack棧

3. 彈出放入result棧

4. stack棧壓入stack的左子節點右子節點

5. 知道stack為空依次彈出result為後序

 1  public void printAfter(TreeNode node){
 2         Stack<TreeNode> stack = new Stack<>();
 3         Stack<TreeNode> result = new Stack<>();
 4         stack.push(node);
 5         while (!stack.isEmpty()){
 6             TreeNode pop = stack.pop();
 7             result.push(pop);
 8             if (pop.left != null){
 9                 stack.push(pop.left);
10             }
11             if (pop.right != null){
12                 stack.push(pop.right);
13             }
14         }
15         while (!result.isEmpty()){
16             TreeNode pop = result.pop();
17             System.out.print(pop.value+" ");
18         }
19 
20     }

寬度優先遍歷

使用一個佇列,先放入頭節點彈出列印然後先放左後放右

 1  public int width(TreeNode node){
 2         Queue<TreeNode> queue = new LinkedList<>();
 3         queue.offer(node);
 4         TreeNode nodeCurEnd = node;
 5         TreeNode nextEnd = null;
 6         int max = Integer.MIN_VALUE;
 7         int curLevelNodes = 0;
 8         while (!queue.isEmpty()){
 9             TreeNode poll = queue.poll();
10             curLevelNodes++;
11             if (poll.left != null){
12                 queue.offer(poll.left);
13                 nextEnd = poll.left;
14             }
15             if (poll.right != null){
16                 queue.offer(poll.right);
17                 nextEnd = poll.right;
18             }
19             if (poll == nodeCurEnd){
20                 max = Math.max(max,curLevelNodes);
21                 curLevelNodes = 0;
22                 nodeCurEnd = nextEnd;
23             }
24         }
25         return Math.max(max,curLevelNodes);
26     }