題目：不分行從上到下打印二叉樹停巷。從上到下打印出二叉樹的每個節(jié)點耍攘，同一層的節(jié)點按照從左到右的順序打印。
思路：就是一個廣度優(yōu)先遍歷畔勤，先把同一層的節(jié)點存儲到隊列中蕾各，然后按照順序取出并且打印，接著把對應(yīng)的左右子節(jié)點加入到隊尾式曲，依次進(jìn)行。
解決方案：

public class Question32 {
    static class BinaryTreeNode{
        double value;
        BinaryTreeNode left;
        BinaryTreeNode right;
        public BinaryTreeNode(int value){
            this.value = value;
        }
    }
    public static void printFromTopToBottom(BinaryTreeNode treeRoot){
        if (treeRoot == null) return;
        Queue<BinaryTreeNode> queue = new LinkedList<>();
        queue.add(treeRoot);
        while (!queue.isEmpty()){
            BinaryTreeNode tmpNode = queue.poll();
            System.out.println(tmpNode.value);
            if (tmpNode.left != null){
                queue.add(tmpNode.left);
            }
            if (tmpNode.right != null){
                queue.add(tmpNode.right);
            }
        }

    }

    public static void main(String[] args) {
        BinaryTreeNode pHead = new BinaryTreeNode(1);
        BinaryTreeNode pAhead = new BinaryTreeNode(3);
        BinaryTreeNode pBhead = new BinaryTreeNode(5);
        BinaryTreeNode pChead = new BinaryTreeNode(7);
        pHead.left = pAhead;
        pHead.right = pBhead;
        pBhead.left = pChead;
        printFromTopToBottom(pHead);
    }
}

題目：分行從上到下打印二叉樹缸榛。從上到下按層打印二叉樹吝羞，同一層的節(jié)點按從左到右的順序打印，每一層打印到一行仔掸。
思路：在如上的基礎(chǔ)上脆贵，加兩個參數(shù)nextLevel和toBePrinted。nextLevel表示下一層的節(jié)點數(shù)起暮，toBePrinted表示在當(dāng)前層中還沒有打印的節(jié)點數(shù)卖氨。
解決方案：

public class Question32 {
    static class BinaryTreeNode{
        double value;
        BinaryTreeNode left;
        BinaryTreeNode right;
        public BinaryTreeNode(int value){
            this.value = value;
        }
    }
   public static void print(BinaryTreeNode treeRoot){
        if (treeRoot == null) return;
        Queue<BinaryTreeNode> queue = new LinkedList<>();
        queue.add(treeRoot);
        int nextLevel = 0;
        int toBePrinted = 1;
        while (!queue.isEmpty()){
            BinaryTreeNode tmpNode = queue.poll();
            System.out.print(tmpNode.value + " ");
            if (tmpNode.left != null){
                queue.add(tmpNode.left);
                ++nextLevel;
            }
            if (tmpNode.right != null){
                queue.add(tmpNode.right);
                ++nextLevel;
            }
            --toBePrinted;
            if (toBePrinted == 0){
                System.out.println("\t");
                toBePrinted = nextLevel;
                nextLevel = 0;
            }
        }
    }

    public static void main(String[] args) {
        BinaryTreeNode pHead = new BinaryTreeNode(1);
        BinaryTreeNode pAhead = new BinaryTreeNode(3);
        BinaryTreeNode pBhead = new BinaryTreeNode(5);
        BinaryTreeNode pChead = new BinaryTreeNode(7);
        pHead.left = pAhead;
        pHead.right = pBhead;
        pBhead.left = pChead;
        print(pHead);
    }
}

題目：之字形打印二叉樹会烙。實現(xiàn)一個函數(shù)按照之字形順序打印二叉樹，即第一層按照從左到右的順序打印筒捺，第二層按照從右到左的順序打印柏腻，其他行以此類推。
思路：構(gòu)建兩個棧系吭，奇數(shù)層五嫂，先保存左子節(jié)點再保存右子節(jié)點；偶數(shù)層肯尺，則先保存右子節(jié)點再保存左子節(jié)點
解決方案：
public class Question32 {
static class BinaryTreeNode{
double value;
BinaryTreeNode left;
BinaryTreeNode right;
public BinaryTreeNode(int value){
this.value = value;
}
}
public static void print3(BinaryTreeNode treeRoot) {
if (treeRoot == null) return;
Stack<BinaryTreeNode> s1 = new Stack<>();
Stack<BinaryTreeNode> s2 = new Stack<>();
s1.add(treeRoot);

    while (!s1.isEmpty() || !s2.isEmpty()) {
        if (!s1.isEmpty()){
            while (!s1.isEmpty()){
                BinaryTreeNode node = s1.pop();
                System.out.print(node.value + " ");
                if (node.left != null){
                    s2.add(node.left);
                }
                if (node.right != null){
                    s2.add(node.right);
                }
            }
            System.out.println();
        }else {
            while (!s2.isEmpty()){
                BinaryTreeNode node = s2.pop();
                System.out.print(node.value + " ");
                if (node.right != null){
                    s1.add(node.right);
                }
                if (node.left != null){
                    s1.add(node.left);
                }
            }
            System.out.println();
        }
    }
}

public static void main(String[] args) {
    BinaryTreeNode pHead = new BinaryTreeNode(1);
    BinaryTreeNode pAhead = new BinaryTreeNode(3);
    BinaryTreeNode pBhead = new BinaryTreeNode(5);
    BinaryTreeNode pChead = new BinaryTreeNode(7);
    pHead.left = pAhead;
    pHead.right = pBhead;
    pBhead.left = pChead;
    print3(pHead);
}

}